How to read resource files in JAR package in Java?

1. Requirements

In a Java project, you need to read files in the resource directory, traverse all files in the specified resource directory, and retain the relative path of the file when reading the file.

2. Problem

When running in IDEA, you can obtain and traverse the specified resources, but after running the Java project into a jar package, you cannot obtain the files in the resource directory.

3. IDEA reads resource resources

After compilation, the resource files are placed in the target directory, and each resource file actually exists on the disk.

3.1, Method 1

Read directly through the absolute path. If file is a directory, you can also recursively traverse the files in the directory through listFiles:

String absolutePath = "资源文件绝对路径";
File file = new File(absolutePath);
if (file.isDirectory()) {
    File[] children = file.listFiles();
}

3.2, Method 2

Reading through relative paths:

String path = "template";    //相对resource路径
File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + path);
if (file.isDirectory()) {
    File[] children = file.listFiles();
}

4. Read resource resources after making them into jar packages

The above two methods cannot read the resource files in the jar package.

After being typed into a jar package, the jar package is a separate file rather than a folder, so the resource file cannot be located through the file path. At this time, the resource files in the jar package can be read through the class loader.

4.1. Read the resource file in the jar package

This method can only read a single file in the jar package, because what is read is the InputStream stream, and the file cannot be retained relative to the resource. path, so the resources in the jar package cannot be traversed.

String path = "/resource相对路径";
InputStream is = this.class.getResourceAsStream(path);
byte[] buff = new byte[1024];
String filePath = "保存文件路径";
String fileName = "保存文件名";
File file = new File(filePath + fileName);
FileUtils.copyInputStreamToFile(is, file);

4.2. Traverse the jar package resource directory

Take copying the resource resource directory as an example and copy the resources in the local and jar packages respectively.

As shown below:

I want to copy all the contents of the template folder in the resource resource directory;

Then save it to C:/Users/ASUS/Desktop/ savePath folder.

4.2.1. Environmental judgment

public static void main(String[] args) throws URISyntaxException {
    // Test为当前类名
    URI uri = Test.class.getProtectionDomain().getCodeSource().getLocation().toURI();
    // tempPath: 文件保存路径
    String tempPath = "C:/Users/ASUS/Desktop/savePath";
    String sourceDir = "template";  //资源文件夹
    if (uri.toString().startsWith("file")) {
        // IDEA运行时，进行资源复制
        copyLocalResourcesFileToTemp(sourceDir + "/", "*", tempPath + "/" + sourceDir);
    } else {
        // 获取jar包所在路径
        String jarPath = uri.toString();
        uri = URI.create(jarPath.substring(jarPath.indexOf("file:"),jarPath.indexOf(".jar") + 4));
        // 打成jar包后，进行资源复制
        Test.copyJarResourcesFileToTemp(uri, tempPath, "BOOT-INF/classes/" + sourceDir);
    }
}

4.2.2. Copy the resource file of the local project

/**
     * 复制本地资源文件到指定目录
     * @param fileRoot      需要复制的资源目录文件夹
     * @param regExpStr     资源文件匹配正则，*表示匹配所有
     * @param tempParent    保存地址
     */
    public static void copyLocalResourcesFileToTemp(String fileRoot, String regExpStr, String tempParent) {
        try {
            ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
            Resource[] resources = resolver.getResources(fileRoot + regExpStr);
            for (Resource resource : resources) {
                File newFile = new File(tempParent, resource.getFilename());
                if (newFile.exists()) {
                    newFile.delete();
                }
                InputStream stream = null;
                try {
                    stream = resource.getInputStream();
                } catch (Exception e) {
                    // 如果resource为文件夹时，会报异常，这里直接忽略这个异常
                }
                if (stream == null) {
                    newFile.mkdirs();
                    copyLocalResourcesFileToTemp(fileRoot + resource.getFilename()  + "/", regExpStr, tempParent + "/" + resource.getFilename());
                } else {
                    if (!newFile.getParentFile().exists()) {
                        newFile.getParentFile().mkdirs();
                    }
                    org.apache.commons.io.FileUtils.copyInputStreamToFile(stream, newFile);
                }
            }
        } catch (Exception e) {
            log.error("failed to copy local source template", e);
        }
    }

4.2.3. Copy the resource files in the jar package

/**
     * 复制jar包中的资源文件到指定目录
     * @param path          jar包所在路径
     * @param tempPath      保存目录
     * @param filePrefix    需要进行复制的资源文件目录：以BOOT-INF/classes/开头
     */
    public static void copyJarResourcesFileToTemp(URI path, String tempPath, String filePrefix) {
        try {
            List<Map.Entry<ZipEntry, InputStream>> collect =
                    readJarFile(new JarFile(path.getPath()), filePrefix).collect(Collectors.toList());
            for (Map.Entry<ZipEntry, InputStream> entry : collect) {
                // 文件相对路径
                String key = entry.getKey().getName();
                // 文件流
                InputStream stream = entry.getValue();
                File newFile = new File(tempPath + key.replaceAll("BOOT-INF/classes", ""));
                if (!newFile.getParentFile().exists()) {
                    newFile.getParentFile().mkdirs();
                }
                org.apache.commons.io.FileUtils.copyInputStreamToFile(stream, newFile);
            }
        } catch (IOException e) {
            log.error("failed to copy jar source template", e);
        }
    }

@SneakyThrows
    public static Stream<Map.Entry<ZipEntry, InputStream>> readJarFile(JarFile jarFile, String prefix) {
        Stream<Map.Entry<ZipEntry, InputStream>> readingStream =
                jarFile.stream().filter(entry -> !entry.isDirectory() && entry.getName().startsWith(prefix))
                        .map(entry -> {
                            try {
                                return new AbstractMap.SimpleEntry<>(entry, jarFile.getInputStream(entry));
                            } catch (IOException e) {
                                return new AbstractMap.SimpleEntry<>(entry, null);
                            }
                        });
        return readingStream.onClose(() -> {
            try {
                jarFile.close();
            } catch (IOException e) {
                log.error("failed to close jarFile", e);
            }
        });
    }

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