How do I load a file from resource folder

Question

My project has the following structure    src main java   src main resources   src test java   src test resources    I have a file in  src test resources test csv and I want to load the file from a unit test in  src test java MyTest java   I have this code which didn t work  It complains  No such file or directory    BufferedReader br   new BufferedReader  new FileReader test csv     I also tried this  InputStream is    InputStream  MyTest class getResourcesAsStream test csv     This also doesn t work  It returns null  I am using Maven to build my project

User · Answer

this getClass   getClassLoader   getResource  filename   getPath

User · Answer

Try   InputStream is   MyTest class getResourceAsStream   test csv      IIRC getResourceAsStream   by default is relative to the class s package   As  Terran noted  don t forget to add the   at the starting of the filename

User · Answer

if you are loading file in static method then ClassLoader classLoader   getClass   getClassLoader    this might give you an error  You can try this e g  file you want to load from resources is resources  gt  gt  Images  gt  gt  Test gif import org springframework core io ClassPathResource  import org springframework core io Resource   Resource resource   new ClassPathResource  quot Images Test gif quot         File file   resource getFile

User · Answer

Now I am illustrating the source code for reading a font from maven created resources directory       scr main resources calibril ttf     Font getCalibriLightFont int fontSize       Font font   null      try          URL fontURL   OneMethod class getResource   calibril ttf            InputStream fontStream   fontURL openStream            font   new Font Font createFont Font TRUETYPE FONT  fontStream  getFamily    Font PLAIN  fontSize           fontStream close         catch IOException   FontFormatException ief           font   new Font  Arial   Font PLAIN  fontSize           ief printStackTrace                 return font      It worked for me and hope that the entire source code will also help you  Enjoy

User · Answer

I got it work on both running jar and in IDE by writing as InputStream schemaStream          ProductUtil class getClassLoader   getResourceAsStream jsonSchemaPath   byte   buffer   new byte schemaStream available     schemaStream read buffer    File tempFile   File createTempFile  quot com package schema testSchema quot    quot json quot    tempFile deleteOnExit    FileOutputStream out   new FileOutputStream tempFile   out write buffer

User · Answer

ClassLoader loader   Thread currentThread   getContextClassLoader    InputStream is   loader getResourceAsStream  test csv      If you use context ClassLoader to find a resource then definitely it will cost application performance

User · Answer

To read the files from  src resources folder then try this   DataSource fds   new FileDataSource getFileHandle  quot images sample jpeg quot      public static File getFileHandle String fileName          return new File YourClassName class getClassLoader   getResource fileName  getFile        in case of non static reference  return new File getClass   getClassLoader   getResource fileName  getFile

User · Answer

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there s a problem with the given filePath   import java io InputStream  import java nio file NoSuchFileException   public class ResourceLoader       private String filePath       public ResourceLoader String filePath                this filePath   filePath           if filePath startsWith                             throw new IllegalArgumentException  Relative paths may not have a leading slash                          public InputStream getResource   throws NoSuchFileException               ClassLoader classLoader   this getClass   getClassLoader             InputStream inputStream   classLoader getResourceAsStream filePath            if inputStream    null                        throw new NoSuchFileException  Resource file not found  Note that the current directory is the source folder                        return inputStream

User · Answer

For java after 1 7   List lt String gt  lines   Files readAllLines Paths get getClass   getResource  test csv   toURI

User · Answer

Here is one quick solution with the use of Guava   import com google common base Charsets  import com google common io Resources   public String readResource final String fileName  Charset charset  throws IOException           return Resources toString Resources getResource fileName   charset       Usage   String fixture   this readResource  filename txt   Charsets UTF 8

User · Answer

Does the code work when not running the Maven-build jar  for example when running from your IDE  If so  make sure the file is actually included in the jar  The resources folder should be included in the pom file  in  lt build gt  lt resources gt

User · Answer

getResource   was working fine with the resources files placed in src main resources only  To get a file which is at the path other than src main resources say src test java you need to create it exlicitly   the following example may help you   import java io BufferedReader  import java io FileReader  import java io IOException  import java net URISyntaxException  import java net URL   public class Main       public static void main String   args  throws URISyntaxException  IOException           URL location   Main class getProtectionDomain   getCodeSource   getLocation            BufferedReader br   new BufferedReader new FileReader location getPath   toString   replace   target classes      src test java youfilename txt

User · Answer

You can use the com google common io Resources getResource to read the url of file and then get the file content using java nio file Files to read the content of file   URL urlPath   Resources getResource  src main resource    List lt String gt  multilineContent  Files readAllLines Paths get urlPath toURI

User · Answer

Try the next   ClassLoader classloader   Thread currentThread   getContextClassLoader    InputStream is   classloader getResourceAsStream  test csv        If the above doesn t work  various projects have been added the following class  ClassLoaderUtil1  code here  2  Here are some examples of how that class is used   src main java com company test YourCallingClass java src main java com opensymphony xwork2 util ClassLoaderUtil java src main resources test csv     java net URL URL url   ClassLoaderUtil getResource  test csv   YourCallingClass class   Path path   Paths get url toURI     List lt String gt  lines   Files readAllLines path  StandardCharsets UTF 8          java io InputStream InputStream inputStream   ClassLoaderUtil getResourceAsStream  test csv   YourCallingClass class   InputStreamReader streamReader   new InputStreamReader inputStream  StandardCharsets UTF 8   BufferedReader reader   new BufferedReader streamReader   for  String line   line   reader readLine       null            Process line       Notes   See it in The Wayback Machine  Also in GitHub

User · Answer

Non spring project  String filePath   Objects requireNonNull getClass   getClassLoader   getResource  quot any json quot    getPath     Stream lt String gt  lines   Files lines Paths get filePath     Or String filePath   Objects requireNonNull getClass   getClassLoader   getResource  quot any json quot    getPath     InputStream in   new FileInputStream filePath    For spring projects  you can also use one line code to get any file under resources folder  File file   ResourceUtils getFile ResourceUtils CLASSPATH URL PREFIX    quot any json quot     String content   new String Files readAllBytes file toPath

User · Answer

My file in the test folder could not be found even though I followed the answers  It got resolved by rebuilding the project  It seems IntelliJ did not recognize the new file automatically  Pretty nasty to find out

User · Answer

I get it to work without any reference to  class  or  ClassLoader    Let s say we have three scenarios with the location of the file  example file  and your working directory  where your app executes  is home mydocuments program projects myapp   a A sub folder descendant to the working directory   myapp res files example file  b A sub folder not descendant to the working directory   projects files example file  b2 Another sub folder not descendant to the working directory   program files example file  c A root folder   home mydocuments files example file  Linux  in Windows replace home  with C    1  Get the right path  a String path    res files example file   b String path       projects files example file  b2 String path          program files example file  c String path     home mydocuments files example file   Basically  if it is a root folder  start the path name with a leading slash  If it is a sub folder  no slash must be before the path name  If the sub folder is not descendant to the working directory you have to cd to it using        This tells the system to go up one folder   2  Create a File object by passing the right path   File file   new File path     3  You are now good to go   BufferedReader br   new BufferedReader new FileReader file

User · Answer

Import the following   import java io IOException  import java io FileNotFoundException  import java io BufferedReader  import java io InputStreamReader  import java io InputStream  import java util ArrayList    The following method returns a file in an ArrayList of Strings   public ArrayList lt String gt  loadFile String filename      ArrayList lt String gt  lines   new ArrayList lt String gt        try       ClassLoader classloader   Thread currentThread   getContextClassLoader        InputStream inputStream   classloader getResourceAsStream filename       InputStreamReader streamReader   new InputStreamReader inputStream  StandardCharsets UTF 8       BufferedReader reader   new BufferedReader streamReader       for  String line   line   reader readLine       null           lines add line             catch FileNotFoundException fnfe          process errors    catch IOException ioe          process errors       return lines

User · Answer

Try Flowing codes on Spring project  ClassPathResource resource   new ClassPathResource  fileName    InputStream inputStream   resource getInputStream      Or on non spring project   ClassLoader classLoader   getClass   getClassLoader     File file   new File classLoader getResource  fileName   getFile      InputStream inputStream   new FileInputStream file

User · Answer

I faced the same issue   The file was not found by a class loader  which means it was not packed into the artifact  jar   You need to build the project  For example  with maven   mvn clean install   So the files you added to resources folder will get into maven build and become available to the application   I would like to keep my answer  it does not explain how to read a file  other answers do explain that   it answers why InputStream or resource was null  Similar answer is here

[java] How do I load a file from resource folder?

Examples related to java

Examples related to file

Examples related to maven