Get resources from a jar file

Question

i want my jar file to access some files from itself. I know how to do this for BufferedImage but this doesn't work for other files. All i want is to extract some zips from my jar. i made a class folder in eclipse, put the zips inside and used

    public File getResFile(String name){

    return new File(getClass().getResource(name).getFile());


}

to get the File instance and extract it. it works fine in eclipse, but as soon as i export it to a jar it says

Exception in thread "main" java.io.FileNotFoundException: file:\C:\Users\DeLL\Desktop\BoxcraftClient\ClientInstaller.jar!\client.bxc (The filename, directory name, or volume label syntax is incorrect)
    at java.util.zip.ZipFile.open(Native Method)
    at java.util.zip.ZipFile.<init>(ZipFile.java:220)
    at java.util.zip.ZipFile.<init>(ZipFile.java:150)
    at java.util.zip.ZipFile.<init>(ZipFile.java:164)
    at Launcher.install(Launcher.java:43)
    at Launcher.main(Launcher.java:33)

Im working to fix this already something like 6 hours and can't find a solution. Please help!

Answer 1

Use one of these methods, from the class Class - getResource(java.lang.String) - getResourceAsStream(java.lang.String)

    this.getClass().getResource(name);
    this.getClass().getResourceAsStream(name);

Warning: By default it loads the file from the location, where the this.class, is found in the package. So if using it from a class org.organisation.project.App, then the file need to be inside the jar in the directory org/organisation/project. In case the file is located in the root, or some other directory, inside the jar, use the /, in from of the file name. Like /data/names.json.

Answer 2

There is a reason why getResource() returns a URL, and not a File, because the resource may not be a file, and since your code is packaged in the Jar file, it's not a file but a zip entry.

The only safe way to read the content of the resource, is as an InputStream, either by calling getResourceAsStream() or by calling openStream() on the returned URL.

Answer 3

first check your class path using java System.out.println("classpath is: " + System.getProperty("java.class.path")); to see if the classpath has your jar file.

And then use the getclass().classloader.getResourceAsStream(name). See if the returned URL is correct. Call the method isFile() on the URL to check if the URL is actually a file. And then call the getFile() method.

Answer 4

Use Spring's PathMatchingResourcePatternResolver;

It will do the trick for both launching the package from an IDE or from the file system:

public List<String> getAllClassesInRunningJar() throws Exception {
    try {
        List<String> list = new ArrayList<String>();

        // Get all the classes inside the package com.my.package:
        // This will do the work for both launching the package from an IDE or from the file system:  
        String scannedPackage = "com.my.package.*";

        // This is spring - org.springframework.core; use these imports:
        //  import org.springframework.core.io.Resource;
        //  import org.springframework.core.io.support.PathMatchingResourcePatternResolver;
        PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
        Resource[] resources = scanner.getResources(scannedPackage.replace(".", "/"));

        for (Resource resource : resources)
            list.add(resource.getURI().toString());
        return list ;
    } catch (Exception e) {
        throw new Exception("Failed to get the classes: " + e.getMessage(), e);
    }
}