列出与 Java 中的模式匹配的目录中的文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2102952/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Listing files in a directory matching a pattern in Java
提问by Omar Kooheji
I'm looking for a way to get a list of files that match a pattern (pref regex) in a given directory.
我正在寻找一种方法来获取与给定目录中的模式（pref regex）匹配的文件列表。
I've found a tutorial online that uses apache's commons-io package with the following code:
我在网上找到了一个使用 apache 的 commons-io 包的教程，代码如下：
Collection getAllFilesThatMatchFilenameExtension(String directoryName, String extension)
{
File directory = new File(directoryName);
return FileUtils.listFiles(directory, new WildcardFileFilter(extension), null);
}
But that just returns a base collection (According to the docsit's a collection of java.io.File). Is there a way to do this that returns a type safe generic collection?
但这只是返回一个基本集合（根据文档，它是一个集合java.io.File）。有没有办法做到这一点，返回一个类型安全的泛型集合？
采纳答案by Kevin
See File#listFiles(FilenameFilter).
请参阅File#listFiles(FilenameFilter)。
File dir = new File(".");
File [] files = dir.listFiles(new FilenameFilter() {
@Override
public boolean accept(File dir, String name) {
return name.endsWith(".xml");
}
});
for (File xmlfile : files) {
System.out.println(xmlfile);
}
回答by jjnguy
The following code will create a list of files based on the accept method of the FileNameFilter.
以下代码将基于 .accept 方法创建文件列表FileNameFilter。
List<File> list = Arrays.asList(dir.listFiles(new FilenameFilter(){
@Override
public boolean accept(File dir, String name) {
return name.endsWith(".exe"); // or something else
}}));
回答by OscarRyz
What about a wrapper around your existing code:
现有代码的包装器怎么样：
public Collection<File> getMatchingFiles( String directory, String extension ) {
return new ArrayList<File>()(
getAllFilesThatMatchFilenameExtension( directory, extension ) );
}
I will throw a warning though. If you can live with that warning, then you're done.
不过我会发出警告。如果你能忍受那个警告，那么你就完成了。
回答by Tarek
Since java 7 you can the java.niopackage to achieve the same result:
从 java 7 开始，您可以使用java.nio包来实现相同的结果：
Path dir = ...;
List<File> files = new ArrayList<>();
try (DirectoryStream<Path> stream = Files.newDirectoryStream(dir, "*.{java,class,jar}")) {
for (Path entry: stream) {
files.add(entry.toFile());
}
return files;
} catch (IOException x) {
throw new RuntimeException(String.format("error reading folder %s: %s",
dir,
x.getMessage()),
x);
}
回答by mr T
Since Java 8 you can use lambdas and achieve shorter code:
从 Java 8 开始，您可以使用 lambdas 并实现更短的代码：
File dir = new File(xmlFilesDirectory);
File[] files = dir.listFiles((d, name) -> name.endsWith(".xml"));
回答by Manash Ranjan Dakua
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;
public class CharCountFromAllFilesInFolder {
public static void main(String[] args)throws IOException {
try{
//C:\Users\MD\Desktop\Test1
System.out.println("Enter Your FilePath:");
Scanner sc = new Scanner(System.in);
Map<Character,Integer> hm = new TreeMap<Character, Integer>();
String s1 = sc.nextLine();
File file = new File(s1);
File[] filearr = file.listFiles();
for (File file2 : filearr) {
System.out.println(file2.getName());
FileReader fr = new FileReader(file2);
BufferedReader br = new BufferedReader(fr);
String s2 = br.readLine();
for (int i = 0; i < s2.length(); i++) {
if(!hm.containsKey(s2.charAt(i))){
hm.put(s2.charAt(i), 1);
}//if
else{
hm.put(s2.charAt(i), hm.get(s2.charAt(i))+1);
}//else
}//for2
System.out.println("The Char Count: "+hm);
}//for1
}//try
catch(Exception e){
System.out.println("Please Give Correct File Path:");
}//catch
}
}
