getClassLoader().getResource("Configuration.xml").getPath() does not work inside a jar

Question

I've created a maven based java project which has resources directory (src/java/resources) and in it some xml files and maven copies them into target/classes/resources (target/classes is my classpath).

To know the content of an xml file, I use:

new FileInputStream(Main.class.getClassLoader().getResource("Configuration.xml").getPath());

Main.class.getClassLoader().getResource("Configuration.xml").getPath() gives me the full path of the xml: "c:\project...\Configuration.xml".

This works well on intellij.

but when I compile and package the project into a jar file I get:

Exception in thread "main" java.io.FileNotFoundException: 
file:\C:\project\target\project-12.50.14-SNAPSHOT-jar-with-dependencies
.jar!\Configuration.xml 

(The filename, directory name, or volume label syntax is incorrect)

That because the path: \C:\project\target\project-12.50.14-SNAPSHOT-jar-with-dependencies .jar!\Configuration.xml cannot be a parameter into FileInputStream().

I cannot replace the getResource() to getResourceAsStream().

I need a way to adjust my jar and resources to work like it work on intellij. I also change the xml resources files content during run-time so that is another problem with keeping them inside a jar.

Can anyone offer me a solution in which I don't have to change me code, and instead change my resources directory structure, or change the classpath, or something else that will keep my code working? Thanks.

"I cannot replace the getResource() to getResourceAsStream()" - why not? Fundamentally, you *can't* use a `FileInputStream` to get at a file embedded within a jar file. You should refactor your code appropriately so that you *can* use `getResourceAsStream`. — Jon Skeet, Aug 27 '15 at 15:22
I cant replace it because this is not my code. I can only consume this class not change it. in addition - I also change connection.xml during run time, so I prefer that this file will site outside the jar (like the directory on intellij "target/class/connection.xml" — Matoy, Aug 27 '15 at 15:30
Ah, I see -- this is a question about building / packaging, not about programming. — Thomas, Aug 27 '15 at 15:34
It appears Intellij puts the directory containing Configuration.xml in the classpath, but when you package the project into a jar, you only have the jar in the classpath. You would need to extract Configuration.xml and make sure it's directory is ahead of the jar on the runtime classpath. — tdrury, Aug 27 '15 at 15:59

Andreas · Answer 1 · 2015-08-27T16:25:52.977

Update: Changed to use File, not Path.

That is because getResource returns a URL, and not all urls are files on disk, i.e. can be converted to a Path. As javadoc says "Returns [...] an empty string if [path part] does not exist".

If it absolutely must be a File or FileInputStream, because some other code you cannot change requires it, then copy the content to a temporary file:

File file = File.createTempFile("Configuration", ".xml");
try {
    try {InputStream in = Main.class.getClassLoader().getResourceAsStream("Configuration.xml")) {
        Files.copy(in, file.toPath()); // requires Java 7
    }
    // use file here, or:
    try {FileInputStream in = new FileInputStream(file)) {
        // use file stream here
    }
} finally {
    file.delete();
}

The above code could be optimized to not copy if resource is a file, but since that's only for development mode, and production mode is always a Jar and will always need the copy, using copy in development mode helps test the code.

getClassLoader().getResource("Configuration.xml").getPath() does not work inside a jar

1 Answers1