Java 7 zip file system provider doesn't seem to accept spaces in URI

Question

I have been testing all possible variations and permutations, but I can\'t seem to construct a FileSystemProvider with the zip/jar scheme for a path (URI) that contains spac

Answer 1

The jar: URIs should have the escaped zip-URI in its scheme-specific part, so your jar: URI is simply wrong - it should rightly be double-escaped, as the jar: scheme is composed of the host URI, !/ and the local path.

However, this escaping is only implied and not expressed by the minimal URL "specification" in JarURLConnection. I agree however with the raised bug in JRE that it should still accept single-escaped, although that could lead to some strange edge-cases not being supported.

As pointed out by tornike and evermean in another answer, the easiest is to do FileSystems.newFileSystem(path, null) - but this does not work when you want to pass and env with say "create"=true.

Instead, create the jar: URI using the component-based constructor:

URI jar = new URI("jar", path.toUri().toString(), null);

This would properly encode the scheme-specific part.

As a JUnit test, which also confirms that this is the escaping used when opening from a Path:

@Test
public void jarWithSpaces() throws Exception {
    Path path = Files.createTempFile("with several spaces", ".zip");
    Files.delete(path);

    // Will fail with FileSystemNotFoundException without env:
    //FileSystems.newFileSystem(path, null);

    // Neither does this work, as it does not double-escape:
    // URI jar = URI.create("jar:" + path.toUri().toASCIIString());                

    URI jar = new URI("jar", path.toUri().toString(), null);
    assertTrue(jar.toASCIIString().contains("with%2520several%2520spaces"));

    Map env = new HashMap<>();
    env.put("create", "true");

    try (FileSystem fs = FileSystems.newFileSystem(jar, env)) {
        URI root = fs.getPath("/").toUri();    
        assertTrue(root.toString().contains("with%2520several%2520spaces"));
    } 
    // Reopen from now-existing Path to check that the URI is
    // escaped in the same way
    try (FileSystem fs = FileSystems.newFileSystem(path, null)) {
        URI root = fs.getPath("/").toUri();
        //System.out.println(root.toASCIIString());
        assertTrue(root.toString().contains("with%2520several%2520spaces"));
    }
}

(I did a similar test with "with\u2301unicode\u263bhere" to check that I did not need to use .toASCIIString())