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How to get a path to a resource in a Java JAR file


Question

I am trying to get a path to a Resource but I have had no luck.

This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents:

ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));

If I do this:

ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());

The result is: java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)

Is there a way to get a path to a resource file?

2013/12/25
1
166
12/25/2013 3:20:43 AM

Accepted Answer

This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.

Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.

2009/06/02
71
6/2/2009 8:39:35 PM

When loading a resource make sure you notice the difference between:

getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path

and

getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning

I guess, this confusion is causing most of problems when loading a resource.


Also, when you're loading an image it's easier to use getResourceAsStream():

BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));

When you really have to load a (non-image) file from a JAR archive, you might try this:

File file = null;
String resource = "/com/myorg/foo.xml";
URL res = getClass().getResource(resource);
if (res.getProtocol().equals("jar")) {
    try {
        InputStream input = getClass().getResourceAsStream(resource);
        file = File.createTempFile("tempfile", ".tmp");
        OutputStream out = new FileOutputStream(file);
        int read;
        byte[] bytes = new byte[1024];

        while ((read = input.read(bytes)) != -1) {
            out.write(bytes, 0, read);
        }
        out.close();
        file.deleteOnExit();
    } catch (IOException ex) {
        Exceptions.printStackTrace(ex);
    }
} else {
    //this will probably work in your IDE, but not from a JAR
    file = new File(res.getFile());
}

if (file != null && !file.exists()) {
    throw new RuntimeException("Error: File " + file + " not found!");
}
2020/06/23

The one line answer is -

String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()

Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm()

References:

getResource(), toExternalForm()

2017/08/30

I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)

try {
    //Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
    //Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
    //Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
    PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.

    //Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
    try {
        PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
    } catch (Exception e) { }

    //Find the last / and cut it off at that location.
    PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
    //If it starts with /, cut it off.
    if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
    //If it starts with file:/, cut that off, too.
    if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
    PROGRAM_DIRECTORY = ""; //Current working directory instead.
}
2011/03/17

if netclient.p is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really file:/path/to/jarfile/bot.jar!/config/netclient.p.

2009/06/02

You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:

src/main/resources/myfile.txt

If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.

2016/02/18

Source: https://stackoverflow.com/questions/941754
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