How can I get a resource Folder from inside my jar File

Question

I have a resources folder package in the root of my project  I  don t  want to load a certain File  If I wanted to load a certain File  I would use class getResourceAsStream and I would be fine   What I actually want to do is to load a  Folder  within the resources folder  loop on the Files inside that Folder and get a Stream to each file and read in the content    Assume that the File names are not determined before runtime    What should I do  Is there a way to get a list of the files inside a Folder in your jar File   Notice that the Jar file with the resources is the same jar file from which the code is being run     Thanks in advance

User · Answer

Try the following  Make the resource path   lt PathRelativeToThisClassFile gt   lt ResourceDirectory gt    E g  if your class path is com abc package MyClass and your resoure files are within src com abc package resources    URL url   MyClass class getResource  resources     if  url    null            error - missing folder   else       File dir   new File url toURI         for  File nextFile   dir listFiles                 Do something with nextFile           You can also use      URL url   MyClass class getResource   com abc package resources

User · Answer

Finally  I found the solution   final String path    sample folder   final File jarFile   new File getClass   getProtectionDomain   getCodeSource   getLocation   getPath      if jarFile isFile          Run with JAR file     final JarFile jar   new JarFile jarFile       final Enumeration lt JarEntry gt  entries   jar entries      gives ALL entries in jar     while entries hasMoreElements              final String name   entries nextElement   getName            if  name startsWith path             filter according to the path             System out println name                       jar close      else      Run with IDE     final URL url   Launcher class getResource       path       if  url    null            try               final File apps   new File url toURI                 for  File app   apps listFiles                      System out println app                           catch  URISyntaxException ex                   never happens                     The second block just work when you run the application on IDE  not with jar file   You can remove it if you don t like that

User · Answer

I know this is many years ago   But just for other people come across this topic  What you could do is to use getResourceAsStream   method with the directory path  and the input Stream will have all the files name from that dir  After that you can concat the dir path with each file name and call getResourceAsStream for each file  in a loop

User · Answer

This link tells you how   The magic is the getResourceAsStream   method    InputStream is    this getClass   getClassLoader   getResourceAsStream  yourpackage mypackage myfile xml

User · Answer

Inside my jar file I had a folder called Upload  this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file  I used the code below   URL inputUrl   getClass   getResource   upload blabla1 txt    File dest1   new File  upload blabla1 txt    FileUtils copyURLToFile inputUrl  dest1    URL inputUrl2   getClass   getResource   upload blabla2 txt    File dest2   new File  upload blabla2 txt    FileUtils copyURLToFile inputUrl2  dest2    URL inputUrl3   getClass   getResource   upload blabla3 txt    File dest3   new File  upload Bblabla3 txt    FileUtils copyURLToFile inputUrl3  dest3

User · Answer

The following code returns the wanted  folder  as Path regardless of if it is inside a jar or not      private Path getFolderPath   throws URISyntaxException  IOException       URI uri   getClass   getClassLoader   getResource  folder   toURI        if   jar  equals uri getScheme             FileSystem fileSystem   FileSystems newFileSystem uri  Collections emptyMap    null         return fileSystem getPath  path to folder inside jar          else         return Paths get uri               Requires java 7

User · Answer

Below code gets  yaml files from a custom resource directory  ClassLoader classLoader   this getClass   getClassLoader                URI uri   classLoader getResource directoryPath  toURI                     if  quot jar quot  equalsIgnoreCase uri getScheme                      Pattern pattern   Pattern compile  quot     quot    quot  classes  quot    directoryPath    quot     yaml  quot                    log debug  quot pattern     quot   pattern pattern                     ApplicationHome home   new ApplicationHome SomeApplication class                   JarFile file   new JarFile home getSource                     Enumeration lt JarEntry gt   jarEntries   file entries                     while jarEntries hasMoreElements                         JarEntry entry   jarEntries nextElement                        Matcher matcher   pattern matcher entry getName                         if matcher find                             InputStream in                                   file getInputStream entry                             work on the stream                                                                     else                    When Spring boot application executed through Non-Jar strategy like through IDE or as a War                       String path   uri getPath                    File   files   new File path  listFiles                                    for File file  files                       if file    null                           try                               InputStream is   new FileInputStream file                                work on stream                                                        catch  Exception e                                log error  quot Exception while parsing file yaml file          quot    file getAbsolutePath    e getMessage                                                    else                          log warn  quot File Object is null while parsing yaml file quot

User · Answer

Simple     use OSGi  In OSGi you can iterate over your Bundle s entries with findEntries and findPaths

User · Answer

As the other answers point out  once the resources are inside a jar file  things get really ugly  In our case  this solution   https   stackoverflow com a 13227570 516188  works very well in the tests  since when the tests are run the code is not packed in a jar file   but doesn t work when the app actually runs normally  So what I ve done is    I hardcode the list of the files in the app  but I have a test which reads the actual list from disk  can do it since that works in tests  and fails if the actual list doesn t match with the list the app returns   That way I have simple code in my app  no tricks   and I m sure I didn t forget to add a new entry in the list thanks to the test

User · Answer

Another solution  you can do it using ResourceLoader like this   import org springframework core io Resource  import org apache commons io FileUtils    Autowire private ResourceLoader resourceLoader        Resource resource   resourceLoader getResource  classpath  path to you dir    File file   resource getFile    Iterator lt File gt  fi   FileUtils iterateFiles file  null  true   while fi hasNext          load fi next

User · Answer

I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar    on both the IDE and on jar  release version    I found java nio file DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar   String fooFolder     foo folder         ClassLoader classLoader   foofClass class getClassLoader    try       uri   classLoader getResource fooFolder  toURI      catch  URISyntaxException e        throw new FooException e getMessage       catch  NullPointerException e       throw new FooException e getMessage        if uri    null       throw new FooException  something is wrong directory or files missing           i want to know if i am inside the jar or working on the IDE   if uri getScheme   contains  jar             jar case        try          URL jar   FooClass class getProtectionDomain   getCodeSource   getLocation              jar toString   begins with file            i want to trim it out            Path jarFile   Paths get jar toString   substring  file   length              FileSystem fs   FileSystems newFileSystem jarFile  null           DirectoryStream lt Path gt  directoryStream   Files newDirectoryStream fs getPath fooFolder            for Path p  directoryStream               InputStream is   FooClass class getResourceAsStream p toString              performFooOverInputStream is               your logic here                        catch IOException e            throw new FooException e getMessage                  else          IDE case        Path path   Paths get uri       try           DirectoryStream lt Path gt  directoryStream   Files newDirectoryStream path           for Path p   directoryStream               InputStream is   new FileInputStream p toFile                 performFooOverInputStream is                   catch  IOException  e            throw new FooException  e getMessage

[java] How can I get a resource "Folder" from inside my jar File?

Examples related to java

Examples related to list

Examples related to resources

Examples related to directory