Reading a resource file from within jar

Solution 1:

Rather than trying to address the resource as a File just ask the ClassLoader to return an InputStream for the resource instead via getResourceAsStream:

try (InputStream in = getClass().getResourceAsStream("/file.txt");
    BufferedReader reader = new BufferedReader(new InputStreamReader(in))) {
    // Use resource
}

As long as the file.txt resource is available on the classpath then this approach will work the same way regardless of whether the file.txt resource is in a classes/ directory or inside a jar.

The URI is not hierarchical occurs because the URI for a resource within a jar file is going to look something like this: file:/example.jar!/file.txt. You cannot read the entries within a jar (a zip file) like it was a plain old File.

This is explained well by the answers to:

• How do I read a resource file from a Java jar file?
• Java Jar file: use resource errors: URI is not hierarchical

Solution 2:

To access a file in a jar you have two options:

• Place the file in directory structure matching your package name (after extracting .jar file, it should be in the same directory as .class file), then access it using getClass().getResourceAsStream("file.txt")

• Place the file at the root (after extracting .jar file, it should be in the root), then access it using Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt")

The first option may not work when jar is used as a plugin.

Solution 3:

I had this problem before and I made fallback way for loading. Basically first way work within .jar file and second way works within eclipse or other IDE.

public class MyClass {

    public static InputStream accessFile() {
        String resource = "my-file-located-in-resources.txt";

        // this is the path within the jar file
        InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
        if (input == null) {
            // this is how we load file within editor (eg eclipse)
            input = MyClass.class.getClassLoader().getResourceAsStream(resource);
        }

        return input;
    }
}

Solution 4:

Up until now (December 2017), this is the only solution I found which works both inside and outside the IDE.

Use PathMatchingResourcePatternResolver

Note: it works also in spring-boot

In this example I'm reading some files located in src/main/resources/my_folder:

try {
    // Get all the files under this inner resource folder: my_folder
    String scannedPackage = "my_folder/*";
    PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
    Resource[] resources = scanner.getResources(scannedPackage);

    if (resources == null || resources.length == 0)
        log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
    else {
        for (Resource resource : resources) {
            log.info(resource.getFilename());
            // Read the file content (I used BufferedReader, but there are other solutions for that):
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
            String line = null;
            while ((line = bufferedReader.readLine()) != null) {
                // ...
                // ...                      
            }
            bufferedReader.close();
        }
    }
} catch (Exception e) {
    throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}

Solution 5:

The problem is that certain third party libraries require file pathnames rather than input streams. Most of the answers don't address this issue.

In this case, one workaround is to copy the resource contents into a temporary file. The following example uses jUnit's TemporaryFolder.

    private List<String> decomposePath(String path){
        List<String> reversed = Lists.newArrayList();
        File currFile = new File(path);
        while(currFile != null){
            reversed.add(currFile.getName());
            currFile = currFile.getParentFile();
        }
        return Lists.reverse(reversed);
    }

    private String writeResourceToFile(String resourceName) throws IOException {
        ClassLoader loader = getClass().getClassLoader();
        InputStream configStream = loader.getResourceAsStream(resourceName);
        List<String> pathComponents = decomposePath(resourceName);
        folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
        File tmpFile = folder.newFile(resourceName);
        Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
        return tmpFile.getAbsolutePath();
    }