java中的接口是类吗
205
2023-07-31
java实现递归文件列表的方法
本文实例讲述了java实现递归文件列表的方法。分享给大家供大家参考。具体如下:
FileListing.java如下:
import java.util.*;
import java.io.*;
/**
* Recursive file listing under a specified directory.
*
* @author javapractices.com
* @author Alex Wong
* @author anonymous user
*/
public final class FileListing {
/**
* Demonstrate use.
*
* @param aArgs - aArgs[0] is the full name of an existing
* directory that can be read.
*/
public static void main(String... aArgs) throws FileNotFoundException {
File startingDirectory= new File(aArgs[0]);
List
//print out all AZwybtaUfile names, in the the order of File.compareTo()
for(File file : files ){
System.out.println(file);
}
}
/**
* Recursively walk a directory tree and return a List of all
* Files found; the List is sorted using File.compareTo().
*
* @param aStartingDir is a validAZwybtaU directory, which can be read.
*/
static public List
File aStartingDir
) throws FileNotFoundException {
validateDirectory(aStartingDir);
List
Collections.sort(result);
return result;
}
// PRIVATE //
static private List
File aStartingDir
) throws FileNotFoundException {
List
File[] filesAndDirs = aStartingDir.listFiles();
List
for(File file : filesDirs) {
result.add(file); //always add, even if directory
if ( ! file.isFile() ) {
//must be a directory
//recursive call!
List
result.addAll(deeperList);
}
}
return result;
}
/**
* Directory is valid if it exists, does not represent a file, and can be read.
*/
static private void validateDirectory (
File aDirectory
) throws FileNotFoundException {
if (aDirectory == null) {
throw new IllegalArgumentException("Directory should not be null.");
}
if (!aDirectory.exists()) {
throw new FileNotFoundException("Directory does not exist: " + aDirectory);
}
if (!aDirectory.isDirectory()) {
throw new IllegalArgumentException("Is not a directory: " + aDirectory);
}
if (!aDirectory.canRead()) {
throw new IllegalArgumentException("Directory cannot be read: " + aDirectory);
}
}
}
希望本文所述对大家的java程序设计有所帮助。
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