How can I get a resource “Folder” from inside my j

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...

Answer 1

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+.

Answer 2

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);

Answer 3

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.

Answer 4

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<Path> 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<Path> directoryStream = Files.newDirectoryStream(path);
        for(Path p : directoryStream){
            InputStream is = new FileInputStream(p.toFile());
            performFooOverInputStream(is);
        }
    } catch (IOException _e) {
        throw new FooException(_e.getMessage());
    }
}

Answer 5

This link tells you how.

The magic is the getResourceAsStream() method :

InputStream is = 
this.getClass().getClassLoader().getResourceAsStream("yourpackage/mypackage/myfile.xml")

Answer 6

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<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
    load(fi.next())
}