下面是代码 来源:forums.devx.com/showthread.php ? t = 153784

import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Enumeration;
import java.util.regex.Pattern;
import java.util.zip.ZipEntry;
import java.util.zip.ZipException;
import java.util.zip.ZipFile;

/**
 * list resources available from the classpath @ *
 */
public class ResourceList{

    /**
     * for all elements of java.class.path get a Collection of resources Pattern
     * pattern = Pattern.compile(".*"); gets all resources
     * 
     * @param pattern
     *            the pattern to match
     * @return the resources in the order they are found
     */
    public static Collection<String> getResources(
        final Pattern pattern){
        final ArrayList<String> retval = new ArrayList<String>();
        final String classPath = System.getProperty("java.class.path", ".");
        final String[] classPathElements = classPath.split(System.getProperty("path.separator"));
        for(final String element : classPathElements){
            retval.addAll(getResources(element, pattern));
        }
        return retval;
    }

    private static Collection<String> getResources(
        final String element,
        final Pattern pattern){
        final ArrayList<String> retval = new ArrayList<String>();
        final File file = new File(element);
        if(file.isDirectory()){
            retval.addAll(getResourcesFromDirectory(file, pattern));
        } else{
            retval.addAll(getResourcesFromJarFile(file, pattern));
        }
        return retval;
    }

    private static Collection<String> getResourcesFromJarFile(
        final File file,
        final Pattern pattern){
        final ArrayList<String> retval = new ArrayList<String>();
        ZipFile zf;
        try{
            zf = new ZipFile(file);
        } catch(final ZipException e){
            throw new Error(e);
        } catch(final IOException e){
            throw new Error(e);
        }
        final Enumeration e = zf.entries();
        while(e.hasMoreElements()){
            final ZipEntry ze = (ZipEntry) e.nextElement();
            final String fileName = ze.getName();
            final boolean accept = pattern.matcher(fileName).matches();
            if(accept){
                retval.add(fileName);
            }
        }
        try{
            zf.close();
        } catch(final IOException e1){
            throw new Error(e1);
        }
        return retval;
    }

    private static Collection<String> getResourcesFromDirectory(
        final File directory,
        final Pattern pattern){
        final ArrayList<String> retval = new ArrayList<String>();
        final File[] fileList = directory.listFiles();
        for(final File file : fileList){
            if(file.isDirectory()){
                retval.addAll(getResourcesFromDirectory(file, pattern));
            } else{
                try{
                    final String fileName = file.getCanonicalPath();
                    final boolean accept = pattern.matcher(fileName).matches();
                    if(accept){
                        retval.add(fileName);
                    }
                } catch(final IOException e){
                    throw new Error(e);
                }
            }
        }
        return retval;
    }

    /**
     * list the resources that match args[0]
     * 
     * @param args
     *            args[0] is the pattern to match, or list all resources if
     *            there are no args
     */
    public static void main(final String[] args){
        Pattern pattern;
        if(args.length < 1){
            pattern = Pattern.compile(".*");
        } else{
            pattern = Pattern.compile(args[0]);
        }
        final Collection<String> list = ResourceList.getResources(pattern);
        for(final String name : list){
            System.out.println(name);
        }
    }
}  

如果你正在使用Spring，请查看PathMatchingResourcePatternResolver

自定义扫描

实现您自己的扫描器。例如:

(评论中提到了此解决方案的局限性)

private List<String> getResourceFiles(String path) throws IOException {
    List<String> filenames = new ArrayList<>();

    try (
            InputStream in = getResourceAsStream(path);
            BufferedReader br = new BufferedReader(new InputStreamReader(in))) {
        String resource;

        while ((resource = br.readLine()) != null) {
            filenames.add(resource);
        }
    }

    return filenames;
}

private InputStream getResourceAsStream(String resource) {
    final InputStream in
            = getContextClassLoader().getResourceAsStream(resource);

    return in == null ? getClass().getResourceAsStream(resource) : in;
}

private ClassLoader getContextClassLoader() {
    return Thread.currentThread().getContextClassLoader();
}

Spring框架

使用Spring框架中的PathMatchingResourcePatternResolver。

Ronmamo反射

对于巨大的CLASSPATH值，其他技术在运行时可能会很慢。一个更快的解决方案是使用ronmamo的Reflections API，它在编译时预编译搜索。

如果你使用apache commonsIO，你可以使用文件系统(可选扩展过滤器):

Collection<File> files = FileUtils.listFiles(new File("directory/"), null, false);

对于资源/类路径:

List<String> files = IOUtils.readLines(MyClass.class.getClassLoader().getResourceAsStream("directory/"), Charsets.UTF_8);

如果您不知道“directoy/”是在文件系统中还是在资源中，您可以添加一个

if (new File("directory/").isDirectory())

or

if (MyClass.class.getClassLoader().getResource("directory/") != null)

在调用和组合使用两者之前…

所以在PathMatchingResourcePatternResolver方面，这是代码中需要的:

@Autowired
ResourcePatternResolver resourceResolver;

public void getResources() {
  resourceResolver.getResources("classpath:config/*.xml");
}

结合了罗伯的回答。

final String resourceDir = "resourceDirectory/";
List<String> files = IOUtils.readLines(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir), Charsets.UTF_8);

for (String f : files) {
  String data = IOUtils.toString(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir + f));
  // ... process data
}

使用反射

获取类路径上的所有内容:

Reflections reflections = new Reflections(null, new ResourcesScanner());
Set<String> resourceList = reflections.getResources(x -> true);

另一个例子-从some.package中获取所有扩展名为.csv的文件:

Reflections reflections = new Reflections("some.package", new ResourcesScanner());
Set<String> resourceList = reflections.getResources(Pattern.compile(".*\\.csv"));

基于上面@rob的信息，我创建了一个实现，我将其发布到公共领域:

private static List<String> getClasspathEntriesByPath(String path) throws IOException {
    InputStream is = Main.class.getClassLoader().getResourceAsStream(path);

    StringBuilder sb = new StringBuilder();
    while (is.available()>0) {
        byte[] buffer = new byte[1024];
        sb.append(new String(buffer, Charset.defaultCharset()));
    }

    return Arrays
            .asList(sb.toString().split("\n"))          // Convert StringBuilder to individual lines
            .stream()                                   // Stream the list
            .filter(line -> line.trim().length()>0)     // Filter out empty lines
            .collect(Collectors.toList());              // Collect remaining lines into a List again
}

虽然我不期望getResourcesAsStream在目录上像那样工作，但它确实做到了，而且工作得很好。

Spring框架的PathMatchingResourcePatternResolver对于这些事情非常棒:

private Resource[] getXMLResources() throws IOException
{
    ClassLoader classLoader = MethodHandles.lookup().getClass().getClassLoader();
    PathMatchingResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(classLoader);

    return resolver.getResources("classpath:x/y/z/*.xml");
}

Maven的依赖:

<dependency>
    <groupId>org.springframework</groupId>
    <artifactId>spring-core</artifactId>
    <version>LATEST</version>
</dependency>

这应该可以工作(如果spring不是一个选项):

public static List<String> getFilenamesForDirnameFromCP(String directoryName) throws URISyntaxException, UnsupportedEncodingException, IOException {
    List<String> filenames = new ArrayList<>();

    URL url = Thread.currentThread().getContextClassLoader().getResource(directoryName);
    if (url != null) {
        if (url.getProtocol().equals("file")) {
            File file = Paths.get(url.toURI()).toFile();
            if (file != null) {
                File[] files = file.listFiles();
                if (files != null) {
                    for (File filename : files) {
                        filenames.add(filename.toString());
                    }
                }
            }
        } else if (url.getProtocol().equals("jar")) {
            String dirname = directoryName + "/";
            String path = url.getPath();
            String jarPath = path.substring(5, path.indexOf("!"));
            try (JarFile jar = new JarFile(URLDecoder.decode(jarPath, StandardCharsets.UTF_8.name()))) {
                Enumeration<JarEntry> entries = jar.entries();
                while (entries.hasMoreElements()) {
                    JarEntry entry = entries.nextElement();
                    String name = entry.getName();
                    if (name.startsWith(dirname) && !dirname.equals(name)) {
                        URL resource = Thread.currentThread().getContextClassLoader().getResource(name);
                        filenames.add(resource.toString());
                    }
                }
            }
        }
    }
    return filenames;
}

使用Spring很简单。无论是一个文件，文件夹，甚至是多个文件，都有可能通过注入来实现。

这个例子演示了插入位于x/y/z文件夹中的多个文件。

import org.springframework.beans.factory.annotation.Value;
import org.springframework.core.io.Resource;
import org.springframework.stereotype.Service;

@Service
public class StackoverflowService {
    @Value("classpath:x/y/z/*")
    private Resource[] resources;

    public List<String> getResourceNames() {
        return Arrays.stream(resources)
                .map(Resource::getFilename)
                .collect(Collectors.toList());
    }
}

它确实适用于文件系统中的资源以及jar中的资源。

我认为您可以利用[Zip文件系统提供程序][1]来实现这一点。当使用文件系统时。newFileSystem，看起来您可以将该ZIP中的对象视为“常规”文件。

在上面的链接文档中:

Specify the configuration options for the zip file system in the java.util.Map object passed to the FileSystems.newFileSystem method. See the [Zip File System Properties][2] topic for information about the provider-specific configuration properties for the zip file system. Once you have an instance of a zip file system, you can invoke the methods of the [java.nio.file.FileSystem][3] and [java.nio.file.Path][4] classes to perform operations such as copying, moving, and renaming files, as well as modifying file attributes.

jdk的文档。在[Java 11状态][5]:

zip文件系统提供程序将zip或JAR文件视为文件系统，并提供操作文件内容的能力。zip文件系统提供程序可以通过[FileSystems]创建。newFileSystem][6]如果已安装。

下面是我利用您的示例资源做的一个虚构的示例。请注意，.zip是.jar，但你可以调整你的代码来使用类路径资源:

设置

cd /tmp
mkdir -p x/y/z
touch x/y/z/{a,b,c}.html
echo 'hello world' > x/y/z/d
zip -r example.zip x

Java

import java.io.IOException;
import java.net.URI;
import java.nio.file.FileSystem;
import java.nio.file.FileSystems;
import java.nio.file.Files;
import java.util.Collections;
import java.util.stream.Collectors;

public class MkobitZipRead {

  public static void main(String[] args) throws IOException {
    final URI uri = URI.create("jar:file:/tmp/example.zip");
    try (
        final FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.emptyMap());
    ) {
      Files.walk(zipfs.getPath("/")).forEach(path -> System.out.println("Files in zip:" + path));
      System.out.println("-----");
      final String manifest = Files.readAllLines(
          zipfs.getPath("x", "y", "z").resolve("d")
      ).stream().collect(Collectors.joining(System.lineSeparator()));
      System.out.println(manifest);
    }
  }

}

输出

Files in zip:/
Files in zip:/x/
Files in zip:/x/y/
Files in zip:/x/y/z/
Files in zip:/x/y/z/c.html
Files in zip:/x/y/z/b.html
Files in zip:/x/y/z/a.html
Files in zip:/x/y/z/d
-----
hello world

这两个答案都不适合我，即使我把我的资源放在资源文件夹里，并遵循上面的答案。真正让人觉得好笑的是:

@Value("file:*/**/resources/**/schema/*.json")
private Resource[] resources;

目前在类路径中列出所有资源的最健壮的机制是与ClassGraph一起使用这种模式，因为它处理了尽可能广泛的类路径规范机制，包括新的JPMS模块系统。(我是ClassGraph的作者。)

List<String> resourceNames;
try (ScanResult scanResult = new ClassGraph().acceptPaths("x/y/z").scan()) {
    resourceNames = scanResult.getAllResources().getNames();
}

我的方法，没有Spring，在单元测试中使用:

URI uri = TestClass.class.getResource("/resources").toURI();
Path myPath = Paths.get(uri);
Stream<Path> walk = Files.walk(myPath, 1);
for (Iterator<Path> it = walk.iterator(); it.hasNext(); ) {
    Path filename = it.next();   
    System.out.println(filename);
}

扩展Luke hutchinson上面的回答，使用他的ClassGraph库，我几乎不费什么力气就能轻松地获得Resource文件夹中所有文件的列表。

假设在资源文件夹中，有一个名为MyImages的文件夹。下面是获取该文件夹中所有文件的URL列表的简单方法:

import io.github.classgraph.ClassGraph;
import io.github.classgraph.ResourceList;
import io.github.classgraph.ScanResult;

public static LinkedList<URL> getURLList (String folder) {
    LinkedList<URL> urlList    = new LinkedList<>();
    ScanResult      scanResult = new ClassGraph().enableAllInfo().scan();
    ResourceList    resources  = scanResult.getAllResources();
    for (URL url : resources.getURLs()) {
        if (url.toString().contains(folder)) {
            urlList.addLast(url);
        }
    }
    return urlList;
}

然后你只需这样做:

LinkedList<URL> myURLFileList = getURLList("MyImages");

这些url可以被加载到流中，或者使用Apache的FileUtils将文件复制到其他地方，就像这样:

String outPath = "/My/Output/Path";
for(URL url : myURLFileList) {
    FileUtils.copyURLToFile(url, new File(outPath, url.getFile()));
}

我认为ClassGraph是一个非常灵巧的库，可以使这样的任务非常简单和容易理解。