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

列出1000多个目录和子目录中文件名的最快方法是什么?

EDIT; The current code I use is:

编辑;我使用的当前代码是:

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);
  }

}

6 个解决方案

#1

This looks fine (Recursively going through the directory) The bottleneck will be all the file i/o you need to do, optimizing your Java will not show any real improvements.

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。

#2

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.

由于这个答案显示在google之上,我正在添加一个用于列出所有文件和目录的java 7 nio解决方案,它在我的系统上花费的时间减少了大约80%。

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();
}

#3

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

唯一的改进是摆脱静态spc_count并将spcs字符串作为参数传递给Process。

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

And when doing recursive call, do

在做递归调用时,做

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

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

这样您就可以从多个线程调用此方法。

#4

Until Java 7 introduces the new java.nio.file classes (like DirectoryStream), I'm afraid what you already have will be the fastest.

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。

#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:

如果您愿意使用第三方库,请查看javaxt-core。它包括一个多线程递归目录搜索,它应该比一次迭代一个目录更快。这里有一些例子:

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

#6

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

我写了一个更简单的代码....试试这个......它将显示每个文件夹,子文件夹和文件......

 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);`

    }

#1