How to get the path of a running JAR file

Question

My code runs inside a JAR file  say foo jar  and I need to know  in the code  in which folder the running foo jar is   So  if foo jar is in C  FOO   I want to get that path no matter what my current working directory is

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Try this   String path   new File     getAbsolutePath

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Have tried several of the solutions up there but none yielded correct results for the  probably special  case that the runnable jar has been exported with  Packaging external libraries  in Eclipse  For some reason all solutions based on the ProtectionDomain do result in null in that case   From combining some solutions above I managed to achieve the following working code   String surroundingJar   null      gets the path to the jar file if it exists  or the  bin  directory if calling from Eclipse String jarDir   new File ClassLoader getSystemClassLoader   getResource      getPath    getAbsolutePath        gets the  bin  directory if calling from eclipse or the name of the  jar file alone  without its path  String jarFileFromSys   System getProperty  java class path   split      0       If both are equal that means it is running from an IDE like Eclipse if  jarFileFromSys equals jarDir         System out println  RUNNING FROM IDE            The path to the jar is the  bin  directory in that case because there is no actual  jar file      surroundingJar   jarDir    else          Combining the path and the name of the  jar file to achieve the final result     surroundingJar   jarDir   jarFileFromSys substring 1      System out println  JAR File      surroundingJar

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Actually here is a better version - the old one failed if a folder name had a space in it     private String getJarFolder            get name and path     String name   getClass   getName   replace                name   getClass   getResource       name     class   toString           remove junk     name   name substring 0  name indexOf   jar         name   name substring name lastIndexOf     -1  name lastIndexOf      1  replace                   remove escape characters     String s           for  int k 0  k lt name length    k            s    name charAt k         if  name charAt k          k    2               replace     with system separator char     return s replace      File separatorChar         As for failing with applets  you wouldn t usually have access to local files anyway  I don t know much about JWS but to handle local files might it not be possible to download the app

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Actually here is a better version - the old one failed if a folder name had a space in it     private String getJarFolder            get name and path     String name   getClass   getName   replace                name   getClass   getResource       name     class   toString           remove junk     name   name substring 0  name indexOf   jar         name   name substring name lastIndexOf     -1  name lastIndexOf      1  replace                   remove escape characters     String s           for  int k 0  k lt name length    k            s    name charAt k         if  name charAt k          k    2               replace     with system separator char     return s replace      File separatorChar         As for failing with applets  you wouldn t usually have access to local files anyway  I don t know much about JWS but to handle local files might it not be possible to download the app

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Use ClassLoader getResource   to find the URL for your current class   For example   package foo   public class Test       public static void main String   args                ClassLoader loader   Test class getClassLoader            System out println loader getResource  foo Test class                This example taken from a similar question    To find the directory  you d then need to take apart the URL manually  See the JarClassLoader tutorial for the format of a jar URL

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If you are really looking for a simple way to get the folder in which your JAR is located you should implement this format of the solution It is easier than what are you are going to find  it is hard to find and many solutions are no longer supported  many others provide the path of the file instead of the actual directory  and this is proven working on version 1 12  new File  quot   quot   getCanonicalPath    Gathering the Input from other answers this is a simple one too  String localPath new File getClass   getProtectionDomain   getCodeSource   getLocation   toURI    getParentFile   getPath    quot    quot     Both will return a String with this format   quot C  Users User Desktop Folder  quot   In a simple and concise line

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Not really sure about the others but in my case it didn t work with a  Runnable jar  and i got it working by fixing codes together from phchen2 answer and another from this link  How to get the path of a running JAR file  The code                  String path new java io File Server class getProtectionDomain                    getCodeSource                    getLocation                    getPath               getAbsolutePath           path path substring 0  path lastIndexOf               path path System getProperty  java class path

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This method  called from code in the archive  returns the folder where the  jar file is  It should work in either Windows or Unix      private String getJarFolder         String name   this getClass   getName   replace                String s   this getClass   getResource       name     class   toString        s   s replace      File separatorChar       s   s substring 0  s indexOf   jar   4       s   s substring s lastIndexOf     -1       return s substring 0  s lastIndexOf File separatorChar  1           Derived from code at  Determine if running from JAR

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return new File MyClass class getProtectionDomain   getCodeSource   getLocation        toURI    getPath      Replace  MyClass  with the name of your class   Obviously  this will do odd things if your class was loaded from a non-file location

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Best solution for me   String path   Test class getProtectionDomain   getCodeSource   getLocation   getPath    String decodedPath   URLDecoder decode path   UTF-8      This should solve the problem with spaces and special characters

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To obtain the File for a given Class  there are two steps    Convert the Class to a URL Convert the URL to a File   It is important to understand both steps  and not conflate them   Once you have the File  you can call getParentFile to get the containing folder  if that is what you need   Step 1  Class to URL  As discussed in other answers  there are two major ways to find a URL relevant to a Class    URL url   Bar class getProtectionDomain   getCodeSource   getLocation    URL url   Bar class getResource Bar class getSimpleName       class      Both have pros and cons   The getProtectionDomain approach yields the base location of the class  e g   the containing JAR file   However  it is possible that the Java runtime s security policy will throw SecurityException when calling getProtectionDomain    so if your application needs to run in a variety of environments  it is best to test in all of them   The getResource approach yields the full URL resource path of the class  from which you will need to perform additional string manipulation  It may be a file  path  but it could also be jar file  or even something nastier like bundleresource   346 fwk2106232034 4 foo Bar class when executing within an OSGi framework  Conversely  the getProtectionDomain approach correctly yields a file  URL even from within OSGi   Note that both getResource     and getResource      failed in my tests  when the class resided within a JAR file  both invocations returned null  So I recommend the  2 invocation shown above instead  as it seems safer   Step 2  URL to File  Either way  once you have a URL  the next step is convert to a File  This is its own challenge  see Kohsuke Kawaguchi s blog post about it for full details  but in short  you can use new File url toURI    as long as the URL is completely well-formed   Lastly  I would highly discourage using URLDecoder  Some characters of the URL    and   in particular  are not valid URL-encoded characters  From the URLDecoder Javadoc      It is assumed that all characters in the encoded string are one of the following   a  through  z    A  through  Z    0  through  9   and  -             and      The character     is allowed but is interpreted as the start of a special escaped sequence                There are two possible ways in which this decoder could deal with illegal strings  It could either leave illegal characters alone or it could throw an IllegalArgumentException  Which approach the decoder takes is left to the implementation    In practice  URLDecoder generally does not throw IllegalArgumentException as threatened above  And if your file path has spaces encoded as  20  this approach may appear to work  However  if your file path has other non-alphameric characters such as   you will have problems with URLDecoder mangling your file path   Working code  To achieve these steps  you might have methods like the following          Gets the base location of the given class      lt p gt     If the class is directly on the file system  e g        path to my package MyClass class   then it will return the base directory     e g    file  path to        lt  p gt      lt p gt     If the class is within a JAR file  e g        path to my-jar jar  my package MyClass class   then it will return the    path to the JAR  e g    file  path to my-jar jar        lt  p gt         param c The class whose location is desired      see FileUtils urlToFile URL  to convert the result to a   link File       public static URL getLocation final Class lt   gt  c        if  c    null  return null     could not load the class         try the easy way first     try           final URL codeSourceLocation               c getProtectionDomain   getCodeSource   getLocation            if  codeSourceLocation    null  return codeSourceLocation            catch  final SecurityException e               NB  Cannot access protection domain            catch  final NullPointerException e               NB  Protection domain or code source is null                NB  The easy way failed  so we try the hard way  We ask for the class        itself as a resource  then strip the class s path from the URL string         leaving the base path          get the class s raw resource path     final URL classResource   c getResource c getSimpleName       class        if  classResource    null  return null     cannot find class resource      final String url   classResource toString        final String suffix   c getCanonicalName   replace               class       if   url endsWith suffix   return null     weird URL         strip the class s path from the URL string     final String base   url substring 0  url length   - suffix length          String path   base          remove the  jar   prefix and      suffix  if present     if  path startsWith  jar     path   path substring 4  path length   - 2        try           return new URL path             catch  final MalformedURLException e            e printStackTrace            return null                   Converts the given   link URL  to its corresponding   link File       lt p gt     This method is similar to calling   code new File url toURI     except that    it also handles  jar file   URLs  returning the path to the JAR file      lt  p gt          param url The URL to convert      return A file path suitable for use with e g    link FileInputStream      throws IllegalArgumentException if the URL does not correspond to a file      public static File urlToFile final URL url        return url    null   null   urlToFile url toString               Converts the given URL string to its corresponding   link File           param url The URL to convert      return A file path suitable for use with e g    link FileInputStream      throws IllegalArgumentException if the URL does not correspond to a file      public static File urlToFile final String url        String path   url      if  path startsWith  jar                  remove  jar   prefix and      suffix         final int index   path indexOf                path   path substring 4  index             try           if  PlatformUtils isWindows    amp  amp  path matches  file  A-Za-z                      path    file      path substring 5                     return new File new URL path  toURI               catch  final MalformedURLException e               NB  URL is not completely well-formed            catch  final URISyntaxException e               NB  URL is not completely well-formed            if  path startsWith  file                  pass through the URL as-is  minus  file   prefix         path   path substring 5           return new File path             throw new IllegalArgumentException  Invalid URL      url       You can find these methods in the SciJava Common library    org scijava util ClassUtils org scijava util FileUtils

User · Answer

Other answers seem to point to the code source which is Jar file location which is not a directory   Use  return new File MyClass class getProtectionDomain   getCodeSource   getLocation   toURI   getPath    getParentFile

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I tried to get the jar running path using   String folder   MyClassName class getProtectionDomain   getCodeSource   getLocation   getPath      c  app java -jar application jar  Running the jar application named  application jar   on Windows in the folder  c  app   the value of the String variable  folder  was   c  app application jar  and I had problems testing for path s correctness    File test   new File folder   if file isDirectory    amp  amp  file canRead        always false     So I tried to define  test  as   String fold  new File folder  getParentFile   getPath   File test   new File fold     to get path in a right format like  c  app  instead of   c  app application jar  and I noticed that it work

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public static String dir   throws URISyntaxException       URI path Main class getProtectionDomain   getCodeSource   getLocation   toURI        String name  Main class getPackage   getName     jar       String path2   path getRawPath        path2 path2 substring 1        if  path2 contains   jar                  path2 path2 replace name                 return path2     Works good on Windows

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This code worked for me to identify if the program is being executed inside a JAR file or IDE   private static boolean isRunningOverJar         try           String pathJar   Application class getResource Application class getSimpleName       class   getFile             if  pathJar toLowerCase   contains   jar                  return true            else               return false                  catch  Exception e            return false            If I need to get the Windows full path of JAR file I am using this method       private static String getPathJar             try               final URI jarUriPath                       Application class getResource Application class getSimpleName       class   toURI                String jarStringPath   jarUriPath toString   replace  jar                     String jarCleanPath    Paths get new URI jarStringPath   toString                 if  jarCleanPath toLowerCase   contains   jar                      return jarCleanPath substring 0  jarCleanPath lastIndexOf   jar     4                 else                   return null                          catch  Exception e                log error  Error getting JAR path    e               return null                    My complete code working with a Spring Boot application using CommandLineRunner implementation  to ensure that the application always be executed within of a console view  Double clicks by mistake in JAR file name   I am using the next code    SpringBootApplication public class Application implements CommandLineRunner       public static void main String   args  throws IOException           Console console   System console             if  console    null  amp  amp   GraphicsEnvironment isHeadless    amp  amp  isRunningOverJar                  Runtime getRuntime   exec new String    cmd     c    start    cmd     k                        java -jar       getPathJar                       else               SpringApplication run Application class  args                         Override     public void run String    args                       Additional code here                          private static boolean isRunningOverJar             try               String pathJar   Application class getResource Application class getSimpleName       class   getFile                 if  pathJar toLowerCase   contains   jar                      return true                else                   return false                          catch  Exception e                return false                       private static String getPathJar             try               final URI jarUriPath                       Application class getResource Application class getSimpleName       class   toURI                String jarStringPath   jarUriPath toString   replace  jar                     String jarCleanPath    Paths get new URI jarStringPath   toString                 if  jarCleanPath toLowerCase   contains   jar                      return jarCleanPath substring 0  jarCleanPath lastIndexOf   jar     4                 else                   return null                          catch  Exception e                return null

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The above methods didn t work for me in my Spring environment  since Spring shades the actual classes into a package called BOOT-INF  thus not the actual location of the running file  I found another way to retrieve the running file through the Permissions object which have been granted to the running file   public static Path getEnclosingDirectory         return Paths get FileUtils class getProtectionDomain   getPermissions                elements   nextElement   getName    getParent

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Use ClassLoader getResource   to find the URL for your current class   For example   package foo   public class Test       public static void main String   args                ClassLoader loader   Test class getClassLoader            System out println loader getResource  foo Test class                This example taken from a similar question    To find the directory  you d then need to take apart the URL manually  See the JarClassLoader tutorial for the format of a jar URL

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I have another way to get the String location of a class   URL path   Thread currentThread   getContextClassLoader   getResource      Path p   Paths get path toURI     String location   p toString      The output String will have the form of   C  Users Administrator new Workspace       The spaces and other characters are handled  and in the form without file    So will be easier to use

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The only solution that works for me on Linux  Mac and Windows   public static String getJarContainingFolder Class aclass  throws Exception     CodeSource codeSource   aclass getProtectionDomain   getCodeSource       File jarFile     if  codeSource getLocation      null        jarFile   new File codeSource getLocation   toURI           else       String path   aclass getResource aclass getSimpleName       class   getPath        String jarFilePath   path substring path indexOf        1  path indexOf            jarFilePath   URLDecoder decode jarFilePath   UTF-8        jarFile   new File jarFilePath         return jarFile getParentFile   getAbsolutePath

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Use ClassLoader getResource   to find the URL for your current class   For example   package foo   public class Test       public static void main String   args                ClassLoader loader   Test class getClassLoader            System out println loader getResource  foo Test class                This example taken from a similar question    To find the directory  you d then need to take apart the URL manually  See the JarClassLoader tutorial for the format of a jar URL

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Mention that it is checked only in Windows but i think it works perfect on other Operating Systems  Linux MacOs Solaris         I had 2  jar files in the same directory   I wanted from the one  jar file to start the other  jar file which is in the same directory   The problem is that when you start it from the cmd the current directory is system32        Warnings     The below seems to work pretty well in all the test i have done even with folder name       57f2g34g87-8 9-09 2        amp    or     amp       it works well  I am using the ProcessBuilder with the below as following          The class from which i called this was the class  Main  String path   getBasePathForClass Main class   String applicationPath   new File path    application jar   getAbsolutePath      System out println  Directory Path is     applicationPath      Your know try catch here   Mention that sometimes it doesn t work for example with folder        57f2g34g87-8 9-09 2        amp      ProcessBuilder builder   new ProcessBuilder  java    -jar   applicationPath   builder redirectErrorStream true   Process process   builder start          code     getBasePathForClass Class lt   gt  classs                   Returns the absolute path of the current directory in which the given        class        file is                  param classs         return The absolute path of the current directory in which the class                file is          author GOXR3PLUS StackOverFlow user    bachden  StackOverFlow user              public static final String getBasePathForClass Class lt   gt  classs                Local variables         File file          String basePath               boolean failed   false              Let s give a first try         try               file   new File classs getProtectionDomain   getCodeSource   getLocation   toURI   getPath                  if  file isFile      file getPath   endsWith   jar      file getPath   endsWith   zip                      basePath   file getParent                  else                   basePath   file getPath                            catch  URISyntaxException ex                failed   true              Logger getLogger classs getName    log Level WARNING                       Cannot firgue out base path for class with way  1      ex                         The above failed          if  failed                try                   file   new File classs getClassLoader   getResource     toURI   getPath                     basePath   file getAbsolutePath                        the below is for testing purposes                       starts with File separator                     String l   local replaceFirst       File separator                                                  catch  URISyntaxException ex                    Logger getLogger classs getName    log Level WARNING                           Cannot firgue out base path for class with way  2      ex                                       fix to run inside eclipse         if  basePath endsWith File separator    lib      basePath endsWith File separator    bin                      basePath endsWith  bin    File separator     basePath endsWith  lib    File separator                 basePath   basePath substring 0  basePath length   - 4                        fix to run inside netbeans         if  basePath endsWith File separator    build    File separator    classes                  basePath   basePath substring 0  basePath length   - 14                        end fix         if   basePath endsWith File separator                 basePath   basePath   File separator                    return basePath

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The getProtectionDomain approach might not work sometimes e g  when you have to find the jar for some of the core java classes  e g in my case StringBuilder class within IBM JDK   however following works seamlessly   public static void main String   args        System out println findSource MyClass class           OR     System out println findSource String class       public static String findSource Class lt   gt  clazz        String resourceToSearch         clazz getName   replace               class       java net URL location   clazz getResource resourceToSearch       String sourcePath   location getPath           Optional  Remove junk     return sourcePath replace  file        replace       resourceToSearch

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the selected answer above is not working if you run your jar by click on it from Gnome desktop environment  not from any script or terminal    Instead  I have fond that the following solution is working everywhere       try           return URLDecoder decode ClassLoader getSystemClassLoader   getResource      getPath     UTF-8          catch  UnsupportedEncodingException e            return

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I had the the same problem and I solved it that way   File currentJavaJarFile   new File Main class getProtectionDomain   getCodeSource   getLocation   getPath        String currentJavaJarFilePath   currentJavaJarFile getAbsolutePath    String currentRootDirectoryPath   currentJavaJarFilePath replace currentJavaJarFile getName           I hope I was of help to you

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Other answers seem to point to the code source which is Jar file location which is not a directory   Use  return new File MyClass class getProtectionDomain   getCodeSource   getLocation   toURI   getPath    getParentFile

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The simplest solution is to pass the path as an argument when running the jar   You can automate this with a shell script   bat in Windows   sh anywhere else    java -jar my-jar jar     I used   to pass the current working directory   UPDATE  You may want to stick the jar file in a sub-directory so users don t accidentally click it  Your code should also check to make sure that the command line arguments have been supplied  and provide a good error message if the arguments are missing

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Have tried several of the solutions up there but none yielded correct results for the  probably special  case that the runnable jar has been exported with  Packaging external libraries  in Eclipse  For some reason all solutions based on the ProtectionDomain do result in null in that case   From combining some solutions above I managed to achieve the following working code   String surroundingJar   null      gets the path to the jar file if it exists  or the  bin  directory if calling from Eclipse String jarDir   new File ClassLoader getSystemClassLoader   getResource      getPath    getAbsolutePath        gets the  bin  directory if calling from eclipse or the name of the  jar file alone  without its path  String jarFileFromSys   System getProperty  java class path   split      0       If both are equal that means it is running from an IDE like Eclipse if  jarFileFromSys equals jarDir         System out println  RUNNING FROM IDE            The path to the jar is the  bin  directory in that case because there is no actual  jar file      surroundingJar   jarDir    else          Combining the path and the name of the  jar file to achieve the final result     surroundingJar   jarDir   jarFileFromSys substring 1      System out println  JAR File      surroundingJar

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You can also use   CodeSource codeSource   YourMainClass class getProtectionDomain   getCodeSource    File jarFile   new File codeSource getLocation   toURI   getPath     String jarDir   jarFile getParentFile   getPath

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I had to mess around a lot before I finally found a working  and short  solution   It is possible that the jarLocation comes with a prefix like file   or jar file   which can be removed by using String substring     URL jarLocationUrl   MyClass class getProtectionDomain   getCodeSource   getLocation    String jarLocation   new File jarLocationUrl toString    getParent

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Here s upgrade to other comments  that seem to me incomplete for the specifics of      using a relative  folder  outside  jar file  in the jar s same   location     String path      YourMainClassName class getProtectionDomain      getCodeSource   getLocation   getPath     path      URLDecoder decode      path        UTF-8     BufferedImage img      ImageIO read      new File           new File path  getParentFile   getPath                File separator             folder             File separator             yourfile jpg

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String path   getClass   getResource     getPath      The path always refers to the resource within the jar file

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I write in Java 7  and test in Windows 7 with Oracle s runtime  and Ubuntu with the open source runtime  This works perfect for those systems   The path for the parent directory of any running jar file  assuming the class calling this code is a direct child of the jar archive itself    try       fooDir   new File this getClass   getClassLoader   getResource     toURI       catch  URISyntaxException e          may be sloppy  but don t really need anything here   fooDirPath   fooDir toString       converts abstract  absolute  path to a String   So  the path of foo jar would be   fooPath   fooDirPath   File separator    foo jar     Again  this wasn t tested on any Mac or older Windows

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return new File MyClass class getProtectionDomain   getCodeSource   getLocation        toURI    getPath      Replace  MyClass  with the name of your class   Obviously  this will do odd things if your class was loaded from a non-file location

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To obtain the File for a given Class  there are two steps    Convert the Class to a URL Convert the URL to a File   It is important to understand both steps  and not conflate them   Once you have the File  you can call getParentFile to get the containing folder  if that is what you need   Step 1  Class to URL  As discussed in other answers  there are two major ways to find a URL relevant to a Class    URL url   Bar class getProtectionDomain   getCodeSource   getLocation    URL url   Bar class getResource Bar class getSimpleName       class      Both have pros and cons   The getProtectionDomain approach yields the base location of the class  e g   the containing JAR file   However  it is possible that the Java runtime s security policy will throw SecurityException when calling getProtectionDomain    so if your application needs to run in a variety of environments  it is best to test in all of them   The getResource approach yields the full URL resource path of the class  from which you will need to perform additional string manipulation  It may be a file  path  but it could also be jar file  or even something nastier like bundleresource   346 fwk2106232034 4 foo Bar class when executing within an OSGi framework  Conversely  the getProtectionDomain approach correctly yields a file  URL even from within OSGi   Note that both getResource     and getResource      failed in my tests  when the class resided within a JAR file  both invocations returned null  So I recommend the  2 invocation shown above instead  as it seems safer   Step 2  URL to File  Either way  once you have a URL  the next step is convert to a File  This is its own challenge  see Kohsuke Kawaguchi s blog post about it for full details  but in short  you can use new File url toURI    as long as the URL is completely well-formed   Lastly  I would highly discourage using URLDecoder  Some characters of the URL    and   in particular  are not valid URL-encoded characters  From the URLDecoder Javadoc      It is assumed that all characters in the encoded string are one of the following   a  through  z    A  through  Z    0  through  9   and  -             and      The character     is allowed but is interpreted as the start of a special escaped sequence                There are two possible ways in which this decoder could deal with illegal strings  It could either leave illegal characters alone or it could throw an IllegalArgumentException  Which approach the decoder takes is left to the implementation    In practice  URLDecoder generally does not throw IllegalArgumentException as threatened above  And if your file path has spaces encoded as  20  this approach may appear to work  However  if your file path has other non-alphameric characters such as   you will have problems with URLDecoder mangling your file path   Working code  To achieve these steps  you might have methods like the following          Gets the base location of the given class      lt p gt     If the class is directly on the file system  e g        path to my package MyClass class   then it will return the base directory     e g    file  path to        lt  p gt      lt p gt     If the class is within a JAR file  e g        path to my-jar jar  my package MyClass class   then it will return the    path to the JAR  e g    file  path to my-jar jar        lt  p gt         param c The class whose location is desired      see FileUtils urlToFile URL  to convert the result to a   link File       public static URL getLocation final Class lt   gt  c        if  c    null  return null     could not load the class         try the easy way first     try           final URL codeSourceLocation               c getProtectionDomain   getCodeSource   getLocation            if  codeSourceLocation    null  return codeSourceLocation            catch  final SecurityException e               NB  Cannot access protection domain            catch  final NullPointerException e               NB  Protection domain or code source is null                NB  The easy way failed  so we try the hard way  We ask for the class        itself as a resource  then strip the class s path from the URL string         leaving the base path          get the class s raw resource path     final URL classResource   c getResource c getSimpleName       class        if  classResource    null  return null     cannot find class resource      final String url   classResource toString        final String suffix   c getCanonicalName   replace               class       if   url endsWith suffix   return null     weird URL         strip the class s path from the URL string     final String base   url substring 0  url length   - suffix length          String path   base          remove the  jar   prefix and      suffix  if present     if  path startsWith  jar     path   path substring 4  path length   - 2        try           return new URL path             catch  final MalformedURLException e            e printStackTrace            return null                   Converts the given   link URL  to its corresponding   link File       lt p gt     This method is similar to calling   code new File url toURI     except that    it also handles  jar file   URLs  returning the path to the JAR file      lt  p gt          param url The URL to convert      return A file path suitable for use with e g    link FileInputStream      throws IllegalArgumentException if the URL does not correspond to a file      public static File urlToFile final URL url        return url    null   null   urlToFile url toString               Converts the given URL string to its corresponding   link File           param url The URL to convert      return A file path suitable for use with e g    link FileInputStream      throws IllegalArgumentException if the URL does not correspond to a file      public static File urlToFile final String url        String path   url      if  path startsWith  jar                  remove  jar   prefix and      suffix         final int index   path indexOf                path   path substring 4  index             try           if  PlatformUtils isWindows    amp  amp  path matches  file  A-Za-z                      path    file      path substring 5                     return new File new URL path  toURI               catch  final MalformedURLException e               NB  URL is not completely well-formed            catch  final URISyntaxException e               NB  URL is not completely well-formed            if  path startsWith  file                  pass through the URL as-is  minus  file   prefix         path   path substring 5           return new File path             throw new IllegalArgumentException  Invalid URL      url       You can find these methods in the SciJava Common library    org scijava util ClassUtils org scijava util FileUtils

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This method  called from code in the archive  returns the folder where the  jar file is  It should work in either Windows or Unix      private String getJarFolder         String name   this getClass   getName   replace                String s   this getClass   getResource       name     class   toString        s   s replace      File separatorChar       s   s substring 0  s indexOf   jar   4       s   s substring s lastIndexOf     -1       return s substring 0  s lastIndexOf File separatorChar  1           Derived from code at  Determine if running from JAR

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Something that is frustrating is that when you are developing in Eclipse MyClass class getProtectionDomain   getCodeSource   getLocation   returns the  bin directory which is great  but when you compile it to a jar  the path includes the  myjarname jar part which gives you illegal file names    To have the code work both in the ide and once it is compiled to a jar  I use the following piece of code   URL applicationRootPathURL   getClass   getProtectionDomain   getCodeSource   getLocation    File applicationRootPath   new File applicationRootPathURL getPath     File myFile  if applicationRootPath isDirectory         myFile   new File applicationRootPath   filename      else      myFile   new File applicationRootPath getParentFile     filename

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the selected answer above is not working if you run your jar by click on it from Gnome desktop environment  not from any script or terminal    Instead  I have fond that the following solution is working everywhere       try           return URLDecoder decode ClassLoader getSystemClassLoader   getResource      getPath     UTF-8          catch  UnsupportedEncodingException e            return

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Try this   String path   new File     getAbsolutePath

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For getting the path of running jar file I have studied the above solutions and tried all methods which exist some difference each other  If these code are running in Eclipse IDE they all should be able to find the path of the file including the indicated class and open or create an indicated file with the found path   But it is tricky  when run the runnable jar file directly or through the command line  it will be failed as the path of jar file gotten from the above methods will give an internal path in the jar file  that is it always gives a path as  rsrc project-name  maybe I should say that it is the package name of the main class file - the indicated class   I can not convert the rsrc     path to an external path  that is when run the jar file outside the Eclipse IDE it can not get the path of jar file   The only possible way for getting the path of running jar file outside Eclipse IDE is   System getProperty  java class path     this code line may return the living path  including the file name  of the running jar file  note that the return path is not the working directory   as the java document and some people said that it will return the paths of all class files in the same directory  but as my tests if in the same directory include many jar files  it only return the path of running jar  about the multiple paths issue indeed it happened in the Eclipse

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The getProtectionDomain approach might not work sometimes e g  when you have to find the jar for some of the core java classes  e g in my case StringBuilder class within IBM JDK   however following works seamlessly   public static void main String   args        System out println findSource MyClass class           OR     System out println findSource String class       public static String findSource Class lt   gt  clazz        String resourceToSearch         clazz getName   replace               class       java net URL location   clazz getResource resourceToSearch       String sourcePath   location getPath           Optional  Remove junk     return sourcePath replace  file        replace       resourceToSearch

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This code worked for me to identify if the program is being executed inside a JAR file or IDE   private static boolean isRunningOverJar         try           String pathJar   Application class getResource Application class getSimpleName       class   getFile             if  pathJar toLowerCase   contains   jar                  return true            else               return false                  catch  Exception e            return false            If I need to get the Windows full path of JAR file I am using this method       private static String getPathJar             try               final URI jarUriPath                       Application class getResource Application class getSimpleName       class   toURI                String jarStringPath   jarUriPath toString   replace  jar                     String jarCleanPath    Paths get new URI jarStringPath   toString                 if  jarCleanPath toLowerCase   contains   jar                      return jarCleanPath substring 0  jarCleanPath lastIndexOf   jar     4                 else                   return null                          catch  Exception e                log error  Error getting JAR path    e               return null                    My complete code working with a Spring Boot application using CommandLineRunner implementation  to ensure that the application always be executed within of a console view  Double clicks by mistake in JAR file name   I am using the next code    SpringBootApplication public class Application implements CommandLineRunner       public static void main String   args  throws IOException           Console console   System console             if  console    null  amp  amp   GraphicsEnvironment isHeadless    amp  amp  isRunningOverJar                  Runtime getRuntime   exec new String    cmd     c    start    cmd     k                        java -jar       getPathJar                       else               SpringApplication run Application class  args                         Override     public void run String    args                       Additional code here                          private static boolean isRunningOverJar             try               String pathJar   Application class getResource Application class getSimpleName       class   getFile                 if  pathJar toLowerCase   contains   jar                      return true                else                   return false                          catch  Exception e                return false                       private static String getPathJar             try               final URI jarUriPath                       Application class getResource Application class getSimpleName       class   toURI                String jarStringPath   jarUriPath toString   replace  jar                     String jarCleanPath    Paths get new URI jarStringPath   toString                 if  jarCleanPath toLowerCase   contains   jar                      return jarCleanPath substring 0  jarCleanPath lastIndexOf   jar     4                 else                   return null                          catch  Exception e                return null

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Use ClassLoader getResource   to find the URL for your current class   For example   package foo   public class Test       public static void main String   args                ClassLoader loader   Test class getClassLoader            System out println loader getResource  foo Test class                This example taken from a similar question    To find the directory  you d then need to take apart the URL manually  See the JarClassLoader tutorial for the format of a jar URL

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I m surprised to see that none recently proposed to use Path  Here follows a citation   The Path class includes various methods that can be used to obtain information about the path  access elements of the path  convert the path to other forms  or extract portions of a path   Thus  a good alternative is to get the Path objest as   Path path   Paths get Test class getProtectionDomain   getCodeSource   getLocation   toURI

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If you are really looking for a simple way to get the folder in which your JAR is located you should implement this format of the solution It is easier than what are you are going to find  it is hard to find and many solutions are no longer supported  many others provide the path of the file instead of the actual directory  and this is proven working on version 1 12  new File  quot   quot   getCanonicalPath    Gathering the Input from other answers this is a simple one too  String localPath new File getClass   getProtectionDomain   getCodeSource   getLocation   toURI    getParentFile   getPath    quot    quot     Both will return a String with this format   quot C  Users User Desktop Folder  quot   In a simple and concise line

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For getting the path of running jar file I have studied the above solutions and tried all methods which exist some difference each other  If these code are running in Eclipse IDE they all should be able to find the path of the file including the indicated class and open or create an indicated file with the found path   But it is tricky  when run the runnable jar file directly or through the command line  it will be failed as the path of jar file gotten from the above methods will give an internal path in the jar file  that is it always gives a path as  rsrc project-name  maybe I should say that it is the package name of the main class file - the indicated class   I can not convert the rsrc     path to an external path  that is when run the jar file outside the Eclipse IDE it can not get the path of jar file   The only possible way for getting the path of running jar file outside Eclipse IDE is   System getProperty  java class path     this code line may return the living path  including the file name  of the running jar file  note that the return path is not the working directory   as the java document and some people said that it will return the paths of all class files in the same directory  but as my tests if in the same directory include many jar files  it only return the path of running jar  about the multiple paths issue indeed it happened in the Eclipse

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The above methods didn t work for me in my Spring environment  since Spring shades the actual classes into a package called BOOT-INF  thus not the actual location of the running file  I found another way to retrieve the running file through the Permissions object which have been granted to the running file   public static Path getEnclosingDirectory         return Paths get FileUtils class getProtectionDomain   getPermissions                elements   nextElement   getName    getParent

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I have another way to get the String location of a class   URL path   Thread currentThread   getContextClassLoader   getResource      Path p   Paths get path toURI     String location   p toString      The output String will have the form of   C  Users Administrator new Workspace       The spaces and other characters are handled  and in the form without file    So will be easier to use

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You can also use   CodeSource codeSource   YourMainClass class getProtectionDomain   getCodeSource    File jarFile   new File codeSource getLocation   toURI   getPath     String jarDir   jarFile getParentFile   getPath

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This code worked for me   private static String getJarPath   throws IOException  URISyntaxException       File f   new File LicensingApp class getProtectionDomain      getLocation   toURI         String jarPath   f getCanonicalPath   toString        String jarDir   jarPath substring  0  jarPath lastIndexOf  File separator         return jarDir

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public static String dir   throws URISyntaxException       URI path Main class getProtectionDomain   getCodeSource   getLocation   toURI        String name  Main class getPackage   getName     jar       String path2   path getRawPath        path2 path2 substring 1        if  path2 contains   jar                  path2 path2 replace name                 return path2     Works good on Windows

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I had the the same problem and I solved it that way   File currentJavaJarFile   new File Main class getProtectionDomain   getCodeSource   getLocation   getPath        String currentJavaJarFilePath   currentJavaJarFile getAbsolutePath    String currentRootDirectoryPath   currentJavaJarFilePath replace currentJavaJarFile getName           I hope I was of help to you

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Not really sure about the others but in my case it didn t work with a  Runnable jar  and i got it working by fixing codes together from phchen2 answer and another from this link  How to get the path of a running JAR file  The code                  String path new java io File Server class getProtectionDomain                    getCodeSource                    getLocation                    getPath               getAbsolutePath           path path substring 0  path lastIndexOf               path path System getProperty  java class path

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Best solution for me   String path   Test class getProtectionDomain   getCodeSource   getLocation   getPath    String decodedPath   URLDecoder decode path   UTF-8      This should solve the problem with spaces and special characters

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Here s upgrade to other comments  that seem to me incomplete for the specifics of      using a relative  folder  outside  jar file  in the jar s same   location     String path      YourMainClassName class getProtectionDomain      getCodeSource   getLocation   getPath     path      URLDecoder decode      path        UTF-8     BufferedImage img      ImageIO read      new File           new File path  getParentFile   getPath                File separator             folder             File separator             yourfile jpg

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Mention that it is checked only in Windows but i think it works perfect on other Operating Systems  Linux MacOs Solaris         I had 2  jar files in the same directory   I wanted from the one  jar file to start the other  jar file which is in the same directory   The problem is that when you start it from the cmd the current directory is system32        Warnings     The below seems to work pretty well in all the test i have done even with folder name       57f2g34g87-8 9-09 2        amp    or     amp       it works well  I am using the ProcessBuilder with the below as following          The class from which i called this was the class  Main  String path   getBasePathForClass Main class   String applicationPath   new File path    application jar   getAbsolutePath      System out println  Directory Path is     applicationPath      Your know try catch here   Mention that sometimes it doesn t work for example with folder        57f2g34g87-8 9-09 2        amp      ProcessBuilder builder   new ProcessBuilder  java    -jar   applicationPath   builder redirectErrorStream true   Process process   builder start          code     getBasePathForClass Class lt   gt  classs                   Returns the absolute path of the current directory in which the given        class        file is                  param classs         return The absolute path of the current directory in which the class                file is          author GOXR3PLUS StackOverFlow user    bachden  StackOverFlow user              public static final String getBasePathForClass Class lt   gt  classs                Local variables         File file          String basePath               boolean failed   false              Let s give a first try         try               file   new File classs getProtectionDomain   getCodeSource   getLocation   toURI   getPath                  if  file isFile      file getPath   endsWith   jar      file getPath   endsWith   zip                      basePath   file getParent                  else                   basePath   file getPath                            catch  URISyntaxException ex                failed   true              Logger getLogger classs getName    log Level WARNING                       Cannot firgue out base path for class with way  1      ex                         The above failed          if  failed                try                   file   new File classs getClassLoader   getResource     toURI   getPath                     basePath   file getAbsolutePath                        the below is for testing purposes                       starts with File separator                     String l   local replaceFirst       File separator                                                  catch  URISyntaxException ex                    Logger getLogger classs getName    log Level WARNING                           Cannot firgue out base path for class with way  2      ex                                       fix to run inside eclipse         if  basePath endsWith File separator    lib      basePath endsWith File separator    bin                      basePath endsWith  bin    File separator     basePath endsWith  lib    File separator                 basePath   basePath substring 0  basePath length   - 4                        fix to run inside netbeans         if  basePath endsWith File separator    build    File separator    classes                  basePath   basePath substring 0  basePath length   - 14                        end fix         if   basePath endsWith File separator                 basePath   basePath   File separator                    return basePath

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The simplest solution is to pass the path as an argument when running the jar   You can automate this with a shell script   bat in Windows   sh anywhere else    java -jar my-jar jar     I used   to pass the current working directory   UPDATE  You may want to stick the jar file in a sub-directory so users don t accidentally click it  Your code should also check to make sure that the command line arguments have been supplied  and provide a good error message if the arguments are missing

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I had to mess around a lot before I finally found a working  and short  solution   It is possible that the jarLocation comes with a prefix like file   or jar file   which can be removed by using String substring     URL jarLocationUrl   MyClass class getProtectionDomain   getCodeSource   getLocation    String jarLocation   new File jarLocationUrl toString    getParent

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I write in Java 7  and test in Windows 7 with Oracle s runtime  and Ubuntu with the open source runtime  This works perfect for those systems   The path for the parent directory of any running jar file  assuming the class calling this code is a direct child of the jar archive itself    try       fooDir   new File this getClass   getClassLoader   getResource     toURI       catch  URISyntaxException e          may be sloppy  but don t really need anything here   fooDirPath   fooDir toString       converts abstract  absolute  path to a String   So  the path of foo jar would be   fooPath   fooDirPath   File separator    foo jar     Again  this wasn t tested on any Mac or older Windows

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I m surprised to see that none recently proposed to use Path  Here follows a citation   The Path class includes various methods that can be used to obtain information about the path  access elements of the path  convert the path to other forms  or extract portions of a path   Thus  a good alternative is to get the Path objest as   Path path   Paths get Test class getProtectionDomain   getCodeSource   getLocation   toURI

User · Answer

String path   getClass   getResource     getPath      The path always refers to the resource within the jar file

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This code worked for me   private static String getJarPath   throws IOException  URISyntaxException       File f   new File LicensingApp class getProtectionDomain      getLocation   toURI         String jarPath   f getCanonicalPath   toString        String jarDir   jarPath substring  0  jarPath lastIndexOf  File separator         return jarDir

User · Answer

return new File MyClass class getProtectionDomain   getCodeSource   getLocation        toURI    getPath      Replace  MyClass  with the name of your class   Obviously  this will do odd things if your class was loaded from a non-file location

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return new File MyClass class getProtectionDomain   getCodeSource   getLocation        toURI    getPath      Replace  MyClass  with the name of your class   Obviously  this will do odd things if your class was loaded from a non-file location

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The only solution that works for me on Linux  Mac and Windows   public static String getJarContainingFolder Class aclass  throws Exception     CodeSource codeSource   aclass getProtectionDomain   getCodeSource       File jarFile     if  codeSource getLocation      null        jarFile   new File codeSource getLocation   toURI           else       String path   aclass getResource aclass getSimpleName       class   getPath        String jarFilePath   path substring path indexOf        1  path indexOf            jarFilePath   URLDecoder decode jarFilePath   UTF-8        jarFile   new File jarFilePath         return jarFile getParentFile   getAbsolutePath

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Something that is frustrating is that when you are developing in Eclipse MyClass class getProtectionDomain   getCodeSource   getLocation   returns the  bin directory which is great  but when you compile it to a jar  the path includes the  myjarname jar part which gives you illegal file names    To have the code work both in the ide and once it is compiled to a jar  I use the following piece of code   URL applicationRootPathURL   getClass   getProtectionDomain   getCodeSource   getLocation    File applicationRootPath   new File applicationRootPathURL getPath     File myFile  if applicationRootPath isDirectory         myFile   new File applicationRootPath   filename      else      myFile   new File applicationRootPath getParentFile     filename

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I tried to get the jar running path using   String folder   MyClassName class getProtectionDomain   getCodeSource   getLocation   getPath      c  app java -jar application jar  Running the jar application named  application jar   on Windows in the folder  c  app   the value of the String variable  folder  was   c  app application jar  and I had problems testing for path s correctness    File test   new File folder   if file isDirectory    amp  amp  file canRead        always false     So I tried to define  test  as   String fold  new File folder  getParentFile   getPath   File test   new File fold     to get path in a right format like  c  app  instead of   c  app application jar  and I noticed that it work

[java] How to get the path of a running JAR file?

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