How do I access a file inside an OSGi bundle?

问题

I am new to OSGi and created an OSGi-bundle which I run in the Apache Felix OSGi-container. There is a file resource contained in the bundle, which I need to pass to a method as a java.io.File. To instantiate a File-object, either an URI in the "file"-scheme or the path as string is necessary. How do I retrieve any of those in a clean way?

I tried using the context.getBundle().getResource("/myfile") (where context is of type org.osgi.framework.BundleContext) which returns the URI bundle://6.0:0/myfile. But this URI can't be converted to a File-instance using the File(URI uri) constructor since it has the "bundle"-scheme.

One could try to construct a path to the location knowing the working directory and exploiting the bundleId of my bundle, but I doubt this is the best practice.

Any ideas?

回答1:

Since the file is inside your bundle, there is no way for you to get to it using a standard File. The URL you get from Bundle.getResource() is the correct way to get to these resources, since the OSGi APIs are intended to also work on systems without an actual file system. I would always try to stick to the OSGi API instead of using framework-specific solutions.

So, if you have control over the method, I would update it to take a URL, or maybe even an InputStream (since you probably just want to read from it). For convenience, you can always provide a helper method that does take a File.

If you don't have control over the method, you will have to write some helper method that takes the URL, streams it out to a file (for instance, File.createTempFile() will probably do the trick.

回答2:

Maybe the API is confusable, but You can access a file inside an OSGI bundle like this:

URL url = context.getBundle().getResource("com/my/weager/impl/test.txt");

// The url maybe like this: bundle://2.0:2/com/my/weager/impl/test.txt
// But this url is not a real file path :(, you could't use it as a file.
// This url should be handled by the specific URLHandlersBundleStreamHandler, 
// you can look up details in BundleRevisionImpl.createURL(int port, String path)
System.out.println(url.toString());

BufferedReader br =new BufferedReader(new InputStreamReader(url.openConnection().getInputStream()));
while(br.ready()){
    System.out.println(br.readLine());
}
br.close();

getResource will find the resource through the whole OSGI container just like OSGI classloader theory.
getEntry will find the resource from local bundle. and the return url could be convert to file but inputStream.
Here is a question same with this: No access to Bundle Resource/File (OSGi) Hope this helping you.

回答3:

What I use is getClassLoader().getResourceAsStream():

InputStream inStream = new java.io.BufferedInputStream(this.getClass().getClassLoader().getResourceAsStream(fileName));

This way the file will be loaded from your resource dir. FileName should contain the path after "src/main/resources".

Full example here:

static public byte[] readFileAsBytes(Class c, String fileName) throws IOException {
    InputStream inStream = new java.io.BufferedInputStream(c.getClassLoader().getResourceAsStream(fileName));
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    int nbytes = 0;
    byte[] buffer = new byte[100000];

    try {
        while ((nbytes = inStream.read(buffer)) != -1) {
            out.write(buffer, 0, nbytes);
        }
        return out.toByteArray();
    } finally {
        if (inStream != null) { 
            inStream.close();
        }
        if (out != null) {
            out.close();
        }
    }
}

来源：https://stackoverflow.com/questions/6474634/how-do-i-access-a-file-inside-an-osgi-bundle

标签

osgi

apache-felix

file-access