11

我想阅读 com.example.resources 包中的一堆文本文件。我可以使用以下代码读取单个文件：

InputStream is = MyObject.class.getResourceAsStream("resources/file1.txt")
InputStreamReader sReader = new InputStreamReader(is);
BefferedReader bReader = new BufferedReader(sReader);
...

有没有办法获取文件列表，然后将每个元素传递给getResourceAsStream？

编辑：根据 ramsinb 的建议，我将代码更改如下：

BufferedReader br = new BufferedReader(new InputStreamReader(MyObject.class.getResourceAsStream("resources")));
String fileName;
while((fileName = br.readLine()) != null){ 
   // access fileName 
}
4

3 回答 3

10

如果您将目录传递给该getResourceAsStream方法，那么它将返回目录中的文件列表（或至少是其中的一个流）。

Thread.currentThread().getContextClassLoader().getResourceAsStream(...)

我故意使用 Thread 来获取资源，因为它将确保我获得父类加载器。这在 Java EE 环境中很重要，但对于您的情况可能并不太重要。

于 2012-12-12T21:06:56.543 回答
3

这个 SO 线程详细讨论了这种技术。下面是一个有用的 Java 方法，它列出了给定资源文件夹中的文件。

/**
   * List directory contents for a resource folder. Not recursive.
   * This is basically a brute-force implementation.
   * Works for regular files and also JARs.
   * 
   * @author Greg Briggs
   * @param clazz Any java class that lives in the same place as the resources you want.
   * @param path Should end with "/", but not start with one.
   * @return Just the name of each member item, not the full paths.
   * @throws URISyntaxException 
   * @throws IOException 
   */
  String[] getResourceListing(Class clazz, String path) throws URISyntaxException, IOException {
      URL dirURL = clazz.getClassLoader().getResource(path);
      if (dirURL != null && dirURL.getProtocol().equals("file")) {
        /* A file path: easy enough */
        return new File(dirURL.toURI()).list();
      } 

      if (dirURL == null) {
        /* 
         * In case of a jar file, we can't actually find a directory.
         * Have to assume the same jar as clazz.
         */
        String me = clazz.getName().replace(".", "/")+".class";
        dirURL = clazz.getClassLoader().getResource(me);
      }

      if (dirURL.getProtocol().equals("jar")) {
        /* A JAR path */
        String jarPath = dirURL.getPath().substring(5, dirURL.getPath().indexOf("!")); //strip out only the JAR file
        JarFile jar = new JarFile(URLDecoder.decode(jarPath, "UTF-8"));
        Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
        Set<String> result = new HashSet<String>(); //avoid duplicates in case it is a subdirectory
        while(entries.hasMoreElements()) {
          String name = entries.nextElement().getName();
          if (name.startsWith(path)) { //filter according to the path
            String entry = name.substring(path.length());
            int checkSubdir = entry.indexOf("/");
            if (checkSubdir >= 0) {
              // if it is a subdirectory, we just return the directory name
              entry = entry.substring(0, checkSubdir);
            }
            result.add(entry);
          }
        }
        return result.toArray(new String[result.size()]);
      } 

      throw new UnsupportedOperationException("Cannot list files for URL "+dirURL);
  }
于 2012-12-12T21:00:05.623 回答
0

我认为这就是你想要的：

String currentDir = new java.io.File(".").toURI().toString();
// AClass = A class in this package
String pathToClass = AClass.class.getResource("/packagename).toString();
String packagePath = (pathToClass.substring(currentDir.length() - 2));

String file;
File folder = new File(packagePath);
File[] filesList= folder.listFiles(); 

for (int i = 0; i < filesList.length; i++) 
{
  if (filesList[i].isFile()) 
  {
    file = filesList[i].getName();
    if (file.endsWith(".txt") || file.endsWith(".TXT"))
    {
      // DO YOUR THING WITH file
    }
  }
}
于 2012-12-12T20:52:57.243 回答