Why is my URI not hierarchical? [duplicate]

Solution 1:

You should be using

getResourceAsStream(...);

when the resource is bundled as a jar/war or any other single file package for that matter.

See the thing is, a jar is a single file (kind of like a zip file) holding lots of files together. From Os's pov, its a single file and if you want to access a part of the file(your image file) you must use it as a stream.

Documentation

Solution 2:

Here is a solution for Eclipse RCP / Plugin developers:

Bundle bundle = Platform.getBundle("resource_from_some_plugin");
URL fileURL = bundle.getEntry("files/test.txt");
File file = null;
try {
   URL resolvedFileURL = FileLocator.toFileURL(fileURL);

   // We need to use the 3-arg constructor of URI in order to properly escape file system chars
   URI resolvedURI = new URI(resolvedFileURL.getProtocol(), resolvedFileURL.getPath(), null);
   File file = new File(resolvedURI);
} catch (URISyntaxException e1) {
    e1.printStackTrace();
} catch (IOException e1) {
    e1.printStackTrace();
}

It's very important to use FileLocator.toFileURL(fileURL) rather than resolve(fileURL) , cause when the plugin is packed into a jar this will cause Eclipse to create an unpacked version in a temporary location so that the object can be accessed using File. For instance, I guess Lars Vogel has an error in his article - http://blog.vogella.com/2010/07/06/reading-resources-from-plugin/

Solution 3:

I face same issue when I was working on a project in my company. First Of All, The URI is not hierarichal Issue is because probably you are using "/" as file separator.

You must remember that "/" is for Windows and from OS to OS it changes, It may be different in Linux. Hence Use File.seperator .

So using

this.getClass().getClassLoader().getResource("res"+File.separator+"secondFolder")

may remove the URI not hierarichal. But Now you may face a Null Pointer Exception. I tried many different ways and then used JarEntries Class to solve it.

File jarFile = new File(this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
        String actualFile = jarFile.getParentFile().getAbsolutePath()+File.separator+"Name_Of_Jar_File.jar";
        System.out.println("jarFile is : "+jarFile.getAbsolutePath());
        System.out.println("actulaFilePath is : "+actualFile);
        final JarFile jar = new JarFile(actualFile);
        final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
        System.out.println("Reading entries in jar file ");
        while(entries.hasMoreElements()) {
            JarEntry jarEntry = entries.nextElement();
            final String name = jarEntry.getName();
            if (name.startsWith("Might Specify a folder name you are searching for")) { //filter according to the path
                System.out.println("file name is "+name);
                System.out.println("is directory : "+jarEntry.isDirectory());
                File scriptsFile  = new File(name);
                System.out.println("file names are : "+scriptsFile.getAbsolutePath());

            }
        }
        jar.close();

You have to specify the jar name here explicitly. So Use this code, this will give you directory and sub directory inside the folder in jar.