How to read files from the sub folder of resource folder with using URI

    How to read files from the sub folder of resource folder.

I have some json file in resources folder like :

src 
    main
        resources 
            jsonData
                d1.json
                d2.json
                d3.json

Now I want to read this in my class which is

src 
    main
        java
            com
                myFile
                    classes 

here is what I am trying.

 File[] fileList = (new File(getClass().getResource("/jaonData").toURI())).listFiles();
    
            for (File file : listOfFiles) {
                if (file.isFile()) {
                   // my operation of Data.
                }
            }

my things are working fine but the problem what I am getting is i don't want to use toURI as it is getting failed.

CodePudding user response：

You're probably not using Spring Boot, so how to read folder from the resolurces files in spring boot, : Getting error while running from Jar won't help you much.

I'll repeat myself from a comment to that question:

Everything inside a JAR file is not a file, and cannot be accessed using File, FileInputStream, etc. There are no official mechanisms to access directories in JAR files.

Fortunately, there is a non-official way, and that uses the fact that you can open a JAR file as a separate file system.

Here's a way that works both with file-based file systems and resources in a JAR file:

private void process() throws IOException {
    Path classBase = getClassBase();
    if (Files.isDirectory(classBase)) {
        process(classBase);
    } else {
        // classBase is the JAR file; open it as a file system
        try (FileSystem fs = FileSystems.newFileSystem(classBase, getClass().getClassLoader())) {
            Path root = fs.getPath("/");
            return loadFromBasePath(root);
        }
    }
} 

private Path getClassBase() {
    ProtectionDomain protectionDomain = getClass().getProtectionDomain();
    CodeSource codeSource = protectionDomain.getCodeSource();
    URL location = codeSource.getLocation();
    try {
        return Paths.get(location.toURI());
    } catch (URISyntaxException e) {
        throw new IllegalStateException(e);
    }
}

private void processRoot(Path root) throws IOException {
    // use root as if it's either the root of the JAR, or target/classes
    // for instance
    Path jsonData = root.resolve("jsonData");
    // Use Files.walk or Files.newDirectoryStream(jsonData)
}

If you don't like using ProtectionDomain, you can use another little trick, that makes use of the fact that every class file can be read as resource:

private Path getClassBase() {
    String resourcePath = '/'   getClass().getName().replace('.', '/')   ".class";
    URL url = getClass().getResource(resourcePath);
    String uriValue = url.toString();
    if (uriValue.endsWith('!'   resourcePath)) {
        // format: jar:<file>!<resourcePath>
        uriValue = uriValue.substring(4, uriValue.length() - resourcePath.length() - 1);
    } else {
        // format: <folder><resourcePath>
        uriValue = uriValue.substring(0, uriValue.length() - resourcePath.length());
    }
    return Paths.get(URI.create(uriValue));
}