Java: Creating a file from a non-hierarchical URI which points to a directory in a jar

Question

I am using reflection to find all the directories in a certain package, and then returning those directories as java.io.Files so that I can extract class files from them. My method for getting all the directories in a package is this:

private static List<File> findDirs(String packageName) throws IOException,
        UnsupportedEncodingException, URISyntaxException {
    List<File> result = new ArrayList<File>();
    ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
    assert classLoader != null;

    String path = packageName.replace('.', '/');
    Enumeration<URL> resources = classLoader.getResources(path);
    while (resources.hasMoreElements()) {
        URL resource = resources.nextElement();
        URI uri = resource.toURI();
        File f = new File(uri);// this line throws the exception
        result.add(f);
    }
    return result;
}

It works perfectly fine when running my application from the source code. However, Once I deploy it,and run it, the method findDirs fails, as the File constructor throws an IllegalArgumentsException"URI is not Hierarchical".

I have debuged the code, and the URI in question which is causing the problem is: jar:file:/C:/Program%20Files/apache-activemq-5.5.1/bin/../lib/App.jar!/com/demo/payload

Whereas I have no problem with the URI when running from source: file:/C:/Marcus/JavaProjects/App/build/prod/classes/com/demo/payload

How can I create a file pointing to the payload directory within the App.jar from a non-herarchical URI such as the one mentioned?

I can get as far as creating a JarFile that points to the App.jar itself using JarURLConnection, and can then get a JarEntry pointing to the directory I'm searching for, but can't seem to convert this entry to a file...

Romain · Accepted Answer · 2012-05-16 10:00:15Z

1

It is wrong to expect you'll be able to treat all stuff you get from #getResources(String) are valid File references. In your run case, it's a reference of a file within your application/library's Jar. You should access these URLs as such: get InputStreams (URL#openStream) on them to read the contents instead of trying to hop through File.

In order to get Files, you'll need to figure out when you're looking at an in-jar URL, then possibly extract the file from the ZIP (because JARs are ZIPs) in order to get a proper, first-class citizen File.

edited May 16, 2012 at 10:00

answered May 16, 2012 at 9:47

Romain

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4 Comments

stryba Over a year ago

-1, as it does not answer the question at all (even though the facts stated are valid)

Marcus Mathioudakis Over a year ago

Agreed, but unfortunately I am working with other peoples code here, and to avoid having to rewrite the entire class I need the findDirs method to return a File pointing to a directory, whether the directory is in a Jar or on the fileSystem, Is there any way to do this?

Romain Over a year ago

@stryba Added notes about getting stuff from Zip. However, it's not something I would recommend doing in a production setup - it's dangerous, classloaders could reference remote data.

stryba Over a year ago

@Romain: agreed not a recommendable solution to the problem

stryba · Accepted Answer · 2012-05-16 09:57:51Z

0

You can use java.util.zip.ZipFile whenever you point to the inside of a jar file. Just load the referenced jar file in a ZipFile and use getEntry or entries of it to retrieve elements within the jar.

answered May 16, 2012 at 9:57

stryba

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2 Comments

Marcus Mathioudakis Over a year ago

As I said at the end of the question I am able to get access to the underlying Zipentry that corresponds to the directory I'm trying to create a file for, but then how can I actually Create a java.io.File instance from this entry?

stryba Over a year ago

The question is whether you need a List<File> rather then something like List<URL> or List<URI> for that matter. With File you won't solve this problem.

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