如何通过Java读取文件夹中的所有文件?这与哪个API无关。


当前回答

我认为这是读取文件夹和子文件夹中的所有文件的好方法

private static void addfiles (File input,ArrayList<File> files)
{
    if(input.isDirectory())
    {
        ArrayList <File> path = new ArrayList<File>(Arrays.asList(input.listFiles()));
        for(int i=0 ; i<path.size();++i)
        {
            if(path.get(i).isDirectory())
            {
                addfiles(path.get(i),files);
            }
            if(path.get(i).isFile())
            {
                files.add(path.get(i));
            }
        }
    }
    if(input.isFile())
    {
        files.add(input);
    }
}

其他回答

public void listFilesForFolder(final File folder) {
    for (final File fileEntry : folder.listFiles()) {
        if (fileEntry.isDirectory()) {
            listFilesForFolder(fileEntry);
        } else {
            System.out.println(fileEntry.getName());
        }
    }
}

final File folder = new File("/home/you/Desktop");
listFilesForFolder(folder);

文件。walk API可从Java 8获得。

try (Stream<Path> paths = Files.walk(Paths.get("/home/you/Desktop"))) {
    paths
        .filter(Files::isRegularFile)
        .forEach(System.out::println);
} 

这个例子使用了API指南中推荐的try-with-resources模式。它确保在任何情况下流都将被关闭。

只需使用“文件”遍历所有“文件”。walkFileTree (java7)

Files.walkFileTree(Paths.get(dir), new SimpleFileVisitor<Path>() {
    @Override
    public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) throws IOException {
        System.out.println("file: " + file);
        return FileVisitResult.CONTINUE;
    }
});
private static final String ROOT_FILE_PATH="/";
File f=new File(ROOT_FILE_PATH);
File[] allSubFiles=f.listFiles();
for (File file : allSubFiles) {
    if(file.isDirectory())
    {
        System.out.println(file.getAbsolutePath()+" is directory");
        //Steps for directory
    }
    else
    {
        System.out.println(file.getAbsolutePath()+" is file");
        //steps for files
    }
}

在Java 7及更高版本中,您可以使用listdir

Path dir = ...;
try (DirectoryStream<Path> stream = Files.newDirectoryStream(dir)) {
    for (Path file: stream) {
        System.out.println(file.getFileName());
    }
} catch (IOException | DirectoryIteratorException x) {
    // IOException can never be thrown by the iteration.
    // In this snippet, it can only be thrown by newDirectoryStream.
    System.err.println(x);
}

您还可以创建一个过滤器,然后将其传递给上面的newDirectoryStream方法

DirectoryStream.Filter<Path> filter = new DirectoryStream.Filter<Path>() {
    public boolean accept(Path file) throws IOException {
        try {
            return (Files.isRegularFile(path));
        } catch (IOException x) {
            // Failed to determine if it's a file.
            System.err.println(x);
            return false;
        }
    }
};

有关其他过滤示例,[参见文档]。(http://docs.oracle.com/javase/tutorial/essential/io/dirs.html#glob)

这将很好地工作:

private static void addfiles(File inputValVal, ArrayList<File> files)
{
  if(inputVal.isDirectory())
  {
    ArrayList <File> path = new ArrayList<File>(Arrays.asList(inputVal.listFiles()));

    for(int i=0; i<path.size(); ++i)
    {
        if(path.get(i).isDirectory())
        {
            addfiles(path.get(i),files);
        }
        if(path.get(i).isFile())
        {
            files.add(path.get(i));
        }
     }

    /*  Optional : if you need to have the counts of all the folders and files you can create 2 global arrays 
        and store the results of the above 2 if loops inside these arrays */
   }

   if(inputVal.isFile())
   {
     files.add(inputVal);
   }

}