列出Java中目录和子目录中的所有文件

Question

What would be the fastest way to list the names of files from 1000+ directories and sub-directories?

EDIT; The current code I use is:

import java.io.File;

public class DirectoryReader {

  static int spc_count=-1;

  static void Process(File aFile) {
    spc_count++;
    String spcs = "";
    for (int i = 0; i < spc_count; i++)
      spcs += " ";
    if(aFile.isFile())
      System.out.println(spcs + "[FILE] " + aFile.getName());
    else if (aFile.isDirectory()) {
      System.out.println(spcs + "[DIR] " + aFile.getName());
      File[] listOfFiles = aFile.listFiles();
      if(listOfFiles!=null) {
        for (int i = 0; i < listOfFiles.length; i++)
          Process(listOfFiles[i]);
      } else {
        System.out.println(spcs + " [ACCESS DENIED]");
      }
    }
    spc_count--;
  }

  public static void main(String[] args) {
    String nam = "D:/";
    File aFile = new File(nam);
    Process(aFile);
  }

}

Answer 1

As this answer shows up on top of google, i'm adding a java 7 nio solution for listing all files and directories, it is takes about 80% less time on my system.

try {
    Path startPath = Paths.get("c:/");
    Files.walkFileTree(startPath, new SimpleFileVisitor<Path>() {
        @Override
        public FileVisitResult preVisitDirectory(Path dir,
                BasicFileAttributes attrs) {
            System.out.println("Dir: " + dir.toString());
            return FileVisitResult.CONTINUE;
        }

        @Override
        public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) {
            System.out.println("File: " + file.toString());    
            return FileVisitResult.CONTINUE;
        }

        @Override
        public FileVisitResult visitFileFailed(Path file, IOException e) {
            return FileVisitResult.CONTINUE;
        }
    });
} catch (IOException e) {
    e.printStackTrace();
}

Answer 2

这看起来很好（递归遍历目录）瓶颈将是你需要做的所有文件i / o，优化你的Java将不会显示任何真正的改进。

Answer 3

The only improvement is to get rid of static spc_count and pass spcs string as a parameter to Process.

public static void main(String[] args) {
  String nam = "D:/";
  File aFile = new File(nam);
  Process("", aFile);
}

And when doing recursive call, do

static void Process( String spcs, File aFile) {
  ...
  Process(spcs + " ", listOfFiles[i]);
  ...
}

This way you can call this method from more than 1 thread.

Answer 4

在Java 7引入新的java.nio.file类（如DirectoryStream ）之前，我担心你已经拥有的将是最快的。

Answer 5

If you're open to using a 3rd party library, check out javaxt-core . It includes a multi-threaded recursive directory search that should be faster than iterating through one directory at a time. There are some examples here:

http://www.javaxt.com/javaxt-core/io/Directory/Recursive_Directory_Search

Answer 6

I have written a much simpler code....Try this... It will show every folder, subfolders and files...

 int Files=0,Directory=0,HiddenFiles=0,HiddenDirectory=0;

 public void listf(String directoryName){

File file=new File(directoryName);

File[] fileList=file.listFiles();

if(fileList!=null){

for(int i=0;i<fileList.length;i++){

if(fileList[i].isHidden()){

if(fileList[i].isFile())

{

   System.out.println(fileList[i]);

HiddenFiles++;

}

else{

  listf(String.valueOf(fileList[i]));

  HiddenDirectory++;

}

}

else if (fileList[i].isFile()) {

//System.out.println(fileList[i]);

Files++;

}

else if(fileList[i].isDirectory()){

Directory++;

listf(String.valueOf(fileList[i]));

}

}

}

}



public void Numbers(){

   System.out.println("Files: "+Files+" HiddenFiles: "+HiddenFiles+"Hidden Directories"+HiddenDirectory+" Directories: "+Directory);`

    }