Java access files in jar causes java.nio.file.FileSystemNotFoundException

Question

While trying to copy some files in my jar file to a temp directory with my java app, the following exception is thrown:

java.nio.file.FileSystemNotFoundException
    at com.sun.nio.zipfs.ZipFileSystemProvider.getFileSystem(ZipFileSystemProvider.java:171)
    at com.sun.nio.zipfs.ZipFileSystemProvider.getPath(ZipFileSystemProvider.java:157)
    at java.nio.file.Paths.get(Unknown Source)
    at com.sora.util.walltoggle.pro.WebViewPresentation.setupTempFiles(WebViewPresentation.java:83)
   ....

and this is a small part of my setupTempFiles(with line numbers):

81. URI uri = getClass().getResource("/webViewPresentation").toURI();
//prints: URI->jar:file:/C:/Users/Tom/Dropbox/WallTogglePro.jar!/webViewPresentation
82. System.out.println("URI->" + uri );
83. Path source = Paths.get(uri);

the webViewPresentation directory resides in the root directory of my jar:

This problem only exits when I package my app as a jar, debugging in Eclipse has no problems. I suspect that this has something to do with this bug but I'm not sure how to correct this problem.

Any helps appreciated

If matters:

I'm on Java 8 build 1.8.0-b132

Windows 7 Ult. x64

Solution 1

A FileSystemNotFoundException means the file system cannot be created automatically; and you have not created it here.

Given your URI, what you should do is split against the !, open the filesystem using the part before it and then get the path from the part after the !:

final Map<String, String> env = new HashMap<>();
final String[] array = uri.toString().split("!");
final FileSystem fs = FileSystems.newFileSystem(URI.create(array[0]), env);
final Path path = fs.getPath(array[1]);

Note that you should .close() your FileSystem once you're done with it.

Solution 2

Accepted answer isn't the best since it doesn't work when you start application in IDE or resource is static and stored in classes! Better solution was proposed at java.nio.file.FileSystemNotFoundException when getting file from resources folder

InputStream in = getClass().getResourceAsStream("/webViewPresentation");
byte[] data = IOUtils.toByteArray(in);

IOUtils is from Apache commons-io.

But if you are already using Spring and want a text file you can change the second line to

StreamUtils.copyToString(in, Charset.defaultCharset());

StreamUtils.copyToByteArray also exists.

Solution 3

This is maybe a hack, but the following worked for me:

URI uri = getClass().getResource("myresourcefile.txt").toURI();

if("jar".equals(uri.getScheme())){
    for (FileSystemProvider provider: FileSystemProvider.installedProviders()) {
        if (provider.getScheme().equalsIgnoreCase("jar")) {
            try {
                provider.getFileSystem(uri);
            } catch (FileSystemNotFoundException e) {
                // in this case we need to initialize it first:
                provider.newFileSystem(uri, Collections.emptyMap());
            }
        }
    }
}
Path source = Paths.get(uri);

This uses the fact that ZipFileSystemProvider internally stores a List of FileSystems that were opened by URI.