从类路径目录中获取资源列表

Question

我正在寻找一种方法来获取给定类路径目录中的所有资源名称列表,类似于方法List<String> getResourceNames (String directoryName).

例如,给定一个路径目录x/y/z包含文件a.html,b.html,c.html和子目录d,getResourceNames("x/y/z")应该返回一个List<String>包含下列字符串:['a.html', 'b.html', 'c.html', 'd'].

它应该适用于文件系统和jar中的资源.

我知道我可以用Files,JarFiles和URLs 写一个快速片段,但我不想重新发明轮子.我的问题是,鉴于现有的公共图书馆,实施最快捷的方式是getResourceNames什么？Spring和Apache Commons堆栈都是可行的.

Answer 1

自定义扫描仪

实现自己的扫描仪.例如:

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PathMatchingResourcePatternResolverSpring Framework.

Ronmamo思考

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

Answer 2

这是代码
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);
        }
    }
}  

如果您使用的是Spring,请查看PathMatchingResourcePatternResolver

Answer 3

使用Google思考:

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获取扩展名为.csv的所有文件:

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

Answer 4

如果您使用apache commonsIO,您可以使用文件系统(可选择使用扩展名过滤器):

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

和资源/类路径:

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

如果你不知道文件系统或资源中是否有"directoy /",你可以添加一个

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

要么

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

在通话之前和两者结合使用......

Answer 5

因此,就PathMatchingResourcePatternResolver而言,这是代码中所需要的:

@Autowired
ResourcePatternResolver resourceResolver;

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

Answer 6

列出类路径中所有资源的最健壮的机制目前是将此模式与 ClassGraph 一起使用，因为它可以处理最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().whitelistPaths("x/y/z").scan()) {
    resourceNames = scanResult.getAllResources().getNames();
}

Answer 7

使用Rob的响应组合.

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
}

Answer 8

该Spring framework的PathMatchingResourcePatternResolver是这些东西真的真棒:

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依赖:

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

Answer 9

这应该有效（如果不能选择 spring）：

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;
}

Answer 10

我的方式，没有 Spring，在单元测试期间使用：

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);
}