如何从Java读取文件夹中的所有文件？

How to read all files in a folder from Java?

如何通过Java读取文件夹中的所有文件？

1
2
3
4
5
6
7
8
9
10
11
12

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);

Field.API可以从Java 8获得。

1
2
3
4
5

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

该示例使用API指南中推荐的Try with Resources模式。它确保无论在什么情况下，流都将关闭。

相关讨论

1
2
3
4
5
6
7
8

File folder = new File("/Users/you/folder/");
File[] listOfFiles = folder.listFiles();

for (File file : listOfFiles) {
if (file.isFile()) {
System.out.println(file.getName());
}
}

相关讨论

在Java 8中，您可以这样做

1
2
3

Files.walk(Paths.get("/path/to/folder"))
.filter(Files::isRegularFile)
.forEach(System.out::println);

它将在排除所有目录的同时打印文件夹中的所有文件。如果需要列表，请执行以下操作：

1
2
3

Files.walk(Paths.get("/path/to/folder"))
.filter(Files::isRegularFile)
.collect(Collectors.toList())

如果您想返回List而不是List，只需映射它：

1
2
3
4

List<File> filesInFolder = Files.walk(Paths.get("/path/to/folder"))
.filter(Files::isRegularFile)
.map(Path::toFile)
.collect(Collectors.toList());

您还需要确保关闭流！否则，您可能会遇到一个异常，告诉您打开的文件太多。阅读此处了解更多信息。

相关讨论

关于使用新Java 8功能的这个主题的所有答案都忽略了关闭流。接受答案中的示例应为：

1
2
3
4
5
6
7

try (Stream<Path> filePathStream=Files.walk(Paths.get("/home/you/Desktop"))) {
filePathStream.forEach(filePath -> {
if (Files.isRegularFile(filePath)) {
System.out.println(filePath);
}
});
}

从Files.walk方法的javadoc中：

The returned stream encapsulates one or more DirectoryStreams. If
timely disposal of file system resources is required, the
try-with-resources construct should be used to ensure that the
stream's close method is invoked after the stream operations are completed.

相关讨论

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

import java.io.File;

public class ReadFilesFromFolder {
public static File folder = new File("C:/Documents and Settings/My Documents/Downloads");
static String temp ="";

public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println("Reading files under the folder"+ folder.getAbsolutePath());
listFilesForFolder(folder);
}

public static void listFilesForFolder(final File folder) {

for (final File fileEntry : folder.listFiles()) {
if (fileEntry.isDirectory()) {
// System.out.println("Reading files under the folder"+folder.getAbsolutePath());
listFilesForFolder(fileEntry);
} else {
if (fileEntry.isFile()) {
temp = fileEntry.getName();
if ((temp.substring(temp.lastIndexOf('.') + 1, temp.length()).toLowerCase()).equals("txt"))
System.out.println("File=" + folder.getAbsolutePath()+"\" + fileEntry.getName());
}

}
}
}
}

相关讨论

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

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
}
}

相关讨论

在Java 7中，您现在可以这样做——http://DOCS.Oracle .COM/JavaSe/Toogs/Value/IO/DRIs。

1
2
3
4
5
6
7
8
9
10

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，然后将其传递到上面的newDirectoryStream方法中。

1
2
3
4
5
6
7
8
9
10
11

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;
}
}
};

其他筛选示例-http://docs.oracle.com/javase/tutorial/essential/io/dirs.html_glob

相关讨论

只需使用EDCOX1×2(Java 7)遍历所有文件

1
2
3
4
5
6
7

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;
}
});

相关讨论

如果需要更多选项，可以使用此函数来填充文件夹中的文件数组列表。选项包括：递归性和要匹配的模式。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38

public static ArrayList<File> listFilesForFolder(final File folder,
final boolean recursivity,
final String patternFileFilter) {

// Inputs
boolean filteredFile = false;

// Ouput
final ArrayList<File> output = new ArrayList<File> ();

// Foreach elements
for (final File fileEntry : folder.listFiles()) {

// If this element is a directory, do it recursivly
if (fileEntry.isDirectory()) {
if (recursivity) {
output.addAll(listFilesForFolder(fileEntry, recursivity, patternFileFilter));
}
}
else {
// If there is no pattern, the file is correct
if (patternFileFilter.length() == 0) {
filteredFile = true;
}
// Otherwise we need to filter by pattern
else {
filteredFile = Pattern.matches(patternFileFilter, fileEntry.getName());
}

// If the file has a name which match with the pattern, then add it to the list
if (filteredFile) {
output.add(fileEntry);
}
}
}

return output;
}

最好的，阿德里安

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

static File mainFolder = new File("Folder");
public static void main(String[] args) {

lf.getFiles(lf.mainFolder);
}
public void getFiles(File f) {
File files[];
if (f.isFile()) {
String name=f.getName();

} else {
files = f.listFiles();
for (int i = 0; i < files.length; i++) {
getFiles(files[i]);
}
}
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

File directory = new File("/user/folder");
File[] myarray;
myarray=new File[10];
myarray=directory.listFiles();
for (int j = 0; j < myarray.length; j++)
{
File path=myarray[j];
FileReader fr = new FileReader(path);
BufferedReader br = new BufferedReader(fr);
String s ="";
while (br.ready()) {
s += br.readLine() +"
";
}
}

相关讨论

如https://stackoverflow.com/a/286001/146745所示，很好地使用了java.io.FileFilter。

1
2
3
4
5
6

File fl = new File(dir);
File[] files = fl.listFiles(new FileFilter() {
public boolean accept(File file) {
return file.isFile();
}
});

相关讨论

我认为这是读取文件夹和子文件夹中所有文件的好方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

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);
}
}

虽然我同意Rich、Orian和其他人使用：

1
2
3
4
5
6
7

final File keysFileFolder = new File(<path>);
File[] fileslist = keysFileFolder.listFiles();

if(fileslist != null)
{
//Do your thing here...
}

出于某种原因，这里的所有示例都使用绝对路径(即，从根目录开始，或者说，对于Windows，使用驱动器号(C:))。

我想补充一下，也可以使用相对路径。因此，如果您的pwd(当前目录/文件夹)是folder1，并且您想要解析folder1/子文件夹，您只需编写(在上面的代码中，而不是在下面的代码中)：

1	final File keysFileFolder = new File("subfolder");

使用Java 1.7递归地在命令行中指定的目录中列出文件的简单示例：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

import java.io.File;

public class List {
public static void main(String[] args) {
for (String f : args) {
listDir(f);
}
}

private static void listDir(String dir) {
File f = new File(dir);
File[] list = f.listFiles();

if (list == null) {
return;
}

for (File entry : list) {
System.out.println(entry.getName());
if (entry.isDirectory()) {
listDir(entry.getAbsolutePath());
}
}
}
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

package com;

import java.io.File;

/**
*
* @author ?Mukesh
*/
public class ListFiles {

static File mainFolder = new File("D:\\Movies");

public static void main(String[] args)
{
ListFiles lf = new ListFiles();
lf.getFiles(lf.mainFolder);

long fileSize = mainFolder.length();
System.out.println("mainFolder size in bytes is:" + fileSize);
System.out.println("File size in KB is :" + (double)fileSize/1024);
System.out.println("File size in MB is :" + (double)fileSize/(1024*1024));
}
public void getFiles(File f){
File files[];
if(f.isFile())
System.out.println(f.getAbsolutePath());
else{
files = f.listFiles();
for (int i = 0; i < files.length; i++) {
getFiles(files[i]);
}
}
}
}

Java 8 EDCOX1(6)是很好的，当你是SOORE，它不会抛出避免Java 8文件。

这里有一个安全的解决方案，虽然不像Java8EDCX1那样优雅，但是6：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

int[] count = {0};
try {
Files.walkFileTree(Paths.get(dir.getPath()), new HashSet<FileVisitOption>(Arrays.asList(FileVisitOption.FOLLOW_LINKS)),
Integer.MAX_VALUE, new SimpleFileVisitor<Path>() {
@Override
public FileVisitResult visitFile(Path file , BasicFileAttributes attrs) throws IOException {
System.out.printf("Visiting file %s
", file);
++count[0];

return FileVisitResult.CONTINUE;
}

@Override
public FileVisitResult visitFileFailed(Path file , IOException e) throws IOException {
System.err.printf("Visiting failed for %s
", file);

return FileVisitResult.SKIP_SUBTREE;
}

@Override
public FileVisitResult preVisitDirectory(Path dir , BasicFileAttributes attrs) throws IOException {
System.out.printf("About to visit directory %s
", dir);
return FileVisitResult.CONTINUE;
}
});
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

package com.commandline.folder;

import java.io.File;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.stream.Stream;

public class FolderReadingDemo {
public static void main(String[] args) {
String str = args[0];
final File folder = new File(str);
// listFilesForFolder(folder);
listFilesForFolder(str);
}

public static void listFilesForFolder(String str) {
try (Stream<Path> paths = Files.walk(Paths.get(str))) {
paths.filter(Files::isRegularFile).forEach(System.out::println);
} catch (Exception e) {
e.printStackTrace();
}
}

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

}

您可以将文件路径放入参数，并使用所有文件路径创建一个列表，而不是手动将其放入列表。然后使用for循环和读卡器。TXT文件示例：

1
2
3
4
5
6
7
8
9
10
11
12

public static void main(String[] args) throws IOException{
File[] files = new File(args[0].replace("\","\\\")).listFiles(new FilenameFilter() { @Override public boolean accept(File dir, String name) { return name.endsWith(".txt"); } });
ArrayList<String> filedir = new ArrayList<String>();
String FILE_TEST = null;
for (i=0; i<files.length; i++){
filedir.add(files[i].toString());
CSV_FILE_TEST=filedir.get(i)

try(Reader testreader = Files.newBufferedReader(Paths.get(FILE_TEST));
){
//write your stuff
}}}

1
2
3
4
5
6
7
8
9
10
11
12

void getFiles(){
String dirPath ="E:/folder_name";
File dir = new File(dirPath);
String[] files = dir.list();
if (files.length == 0) {
System.out.println("The directory is empty");
} else {
for (String aFile : files) {
System.out.println(aFile);
}
}
}

根据获取目录中所有文件的说明。方法Files.walk(path)将通过遍历给定开始文件的根文件树来返回所有文件。

例如，下一个文件树是：

1
2
3
4
5
6
7

\---folder
| file1.txt
| file2.txt
|
\---subfolder
file3.txt
file4.txt

使用java.nio.file.Files.walk(Path)时：

1
2
3

Files.walk(Paths.get("folder"))
.filter(Files::isRegularFile)
.forEach(System.out::println);

给出以下结果：

1
2
3
4

folder\file1.txt
folder\file2.txt
folder\subfolder\file3.txt
folder\subfolder\file4.txt

要仅获取当前目录中的所有文件，请使用java.nio.file.Files.list(Path)：

1
2
3

Files.list(Paths.get("folder"))
.filter(Files::isRegularFile)
.forEach(System.out::println);

结果：

1 2	folder\file1.txt folder\file2.txt

list down files from Test folder present inside class path

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

import java.io.File;
import java.io.IOException;

public class Hello {

public static void main(final String[] args) throws IOException {

System.out.println("List down all the files present on the server directory");
File file1 = new File("/prog/FileTest/src/Test");
File[] files = file1.listFiles();
if (null != files) {
for (int fileIntList = 0; fileIntList < files.length; fileIntList++) {
String ss = files[fileIntList].toString();
if (null != ss && ss.length() > 0) {
System.out.println("File:" + (fileIntList + 1) +" :" + ss.substring(ss.lastIndexOf("\") + 1, ss.length()));
}
}
}

}

}

为了扩展接受的答案，我将文件名存储到arraylist(而不是将其转储到system.out.println)，我创建了一个助手类"myfileutils"，以便其他项目可以导入它：

1
2
3
4
5
6
7
8
9
10
11

class MyFileUtils {
public static void loadFilesForFolder(final File folder, List<String> fileList){
for (final File fileEntry : folder.listFiles()) {
if (fileEntry.isDirectory()) {
loadFilesForFolder(fileEntry, fileList);
} else {
fileList.add( fileEntry.getParent() + File.separator + fileEntry.getName() );
}
}
}
}

我添加了文件名的完整路径。您可以这样使用它：

1
2
3
4
5
6
7
8
9
10

import MyFileUtils;

List<String> fileList = new ArrayList<String>();
final File folder = new File("/home/you/Desktop");
MyFileUtils.loadFilesForFolder(folder, fileList);

// Dump file list values
for (String fileName : fileList){
System.out.println(fileName);
}

arraylist由"value"传递，但该值用于指向生活在JVM堆中的相同arraylist对象。这样，每个递归调用都会将文件名添加到同一个arraylist(我们不会在每个递归调用上创建新的arraylist)。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38

/**
* Function to read all mp3 files from sdcard and store the details in an
* ArrayList
*/

public ArrayList<HashMap<String, String>> getPlayList()
{
ArrayList<HashMap<String, String>> songsList=new ArrayList<>();
File home = new File(MEDIA_PATH);

if (home.listFiles(new FileExtensionFilter()).length > 0) {
for (File file : home.listFiles(new FileExtensionFilter())) {
HashMap<String, String> song = new HashMap<String, String>();
song.put(
"songTitle",
file.getName().substring(0,
(file.getName().length() - 4)));
song.put("songPath", file.getPath());

// Adding each song to SongList
songsList.add(song);
}
}
// return songs list array
return songsList;
}

/**
* Class to filter files which have a .mp3 extension
* */
class FileExtensionFilter implements FilenameFilter
{
@Override
public boolean accept(File dir, String name) {
return (name.endsWith(".mp3") || name.endsWith(".MP3"));
}
}

您可以过滤任何文本文件或任何其他扩展名..只需将其替换为.mp3

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

import java.io.File;
import java.util.ArrayList;
import java.util.List;

public class AvoidNullExp {

public static void main(String[] args) {

List<File> fileList =new ArrayList<>();
final File folder = new File("g:/master");
new AvoidNullExp().listFilesForFolder(folder, fileList);
}

public void listFilesForFolder(final File folder,List<File> fileList) {
File[] filesInFolder = folder.listFiles();
if (filesInFolder != null) {
for (final File fileEntry : filesInFolder) {
if (fileEntry.isDirectory()) {
System.out.println("DIR :"+fileEntry.getName());
listFilesForFolder(fileEntry,fileList);
} else {
System.out.println("FILE :"+fileEntry.getName());
fileList.add(fileEntry);
}
}
}
}

}

这将读取给定路径中的指定文件扩展名文件(同时查找子文件夹)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

public static Map<String,List<File>> getFileNames(String
dirName,Map<String,List<File>> filesContainer,final String fileExt){
String dirPath = dirName;
List<File>files = new ArrayList<>();
Map<String,List<File>> completeFiles = filesContainer;
if(completeFiles == null) {
completeFiles = new HashMap<>();
}
File file = new File(dirName);

FileFilter fileFilter = new FileFilter() {
@Override
public boolean accept(File file) {
boolean acceptFile = false;
if(file.isDirectory()) {
acceptFile = true;
}else if (file.getName().toLowerCase().endsWith(fileExt))
{
acceptFile = true;
}
return acceptFile;
}
};
for(File dirfile : file.listFiles(fileFilter)) {
if(dirfile.isFile() &&
dirfile.getName().toLowerCase().endsWith(fileExt)) {
files.add(dirfile);
}else if(dirfile.isDirectory()) {
if(!files.isEmpty()) {
completeFiles.put(dirPath, files);
}

getFileNames(dirfile.getAbsolutePath(),completeFiles,fileExt);
}
}
if(!files.isEmpty()) {
completeFiles.put(dirPath, files);
}
return completeFiles;
}

相关讨论

上面有很多很好的答案，这里有一种不同的方法：在Maven项目中，默认情况下，您放入Resources文件夹中的所有内容都会复制到Target/Classes文件夹中。查看运行时可用的内容

1
2
3
4
5
6
7
8

ClassLoader contextClassLoader =
Thread.currentThread().getContextClassLoader();
URL resource = contextClassLoader.getResource("");
File file = new File(resource.toURI());
File[] files = file.listFiles();
for (File f : files) {
System.out.println(f.getName());
}

现在要从特定文件夹获取文件，假设您的资源文件夹中有一个名为"res"的文件夹，只需替换：

1	URL resource = contextClassLoader.getResource("res");

如果要在com.companyname包中访问，请执行以下操作：

1	contextClassLoader.getResource("com.companyName");

给定一个basedir，列出它下面的所有文件和目录，迭代地编写。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

public static List<File> listLocalFilesAndDirsAllLevels(File baseDir) {

List<File> collectedFilesAndDirs = new ArrayList<>();
Deque<File> remainingDirs = new ArrayDeque<>();

if(baseDir.exists()) {
remainingDirs.add(baseDir);

while(!remainingDirs.isEmpty()) {
File dir = remainingDirs.removeLast();
List<File> filesInDir = Arrays.asList(dir.listFiles());
for(File fileOrDir : filesInDir) {
collectedFilesAndDirs.add(fileOrDir);
if(fileOrDir.isDirectory()) {
remainingDirs.add(fileOrDir);
}
}
}
}

return collectedFilesAndDirs;
}

这样做很好：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28

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);
}

}

为了防止在listfiles()函数上出现nullpointerException，并递归地从子目录中获取所有文件。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

public void listFilesForFolder(final File folder,List<File> fileList) {
File[] filesInFolder = folder.listFiles();
if (filesInFolder != null) {
for (final File fileEntry : filesInFolder) {
if (fileEntry.isDirectory()) {
listFilesForFolder(fileEntry,fileList);
} else {
fileList.add(fileEntry);
}
}
}
}

List<File> fileList = new List<File>();
final File folder = new File("/home/you/Desktop");
listFilesForFolder(folder);

相关讨论

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

import java.io.File;

public class Test {

public void test1() {
System.out.println("TEST 1");
}

public static void main(String[] args) throws SecurityException, ClassNotFoundException{

File actual = new File("src");
File list[] = actual.listFiles();
for(int i=0; i<list.length; i++){
String substring = list[i].getName().substring(0, list[i].getName().indexOf("."));
if(list[i].isFile() && list[i].getName().contains(".java")){
if(Class.forName(substring).getMethods()[0].getName().contains("main")){
System.out.println("CLASS NAME"+Class.forName(substring).getName());
}

}
}

}
}

只要传递你的文件夹，它就会告诉你方法的主要类。