How to read all files in a folder from Java?
JavaFileIoDirectoryJava Problem Overview
How to read all the files in a folder through Java? It doesn't matter which API.
Java Solutions
Solution 1 - Java
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);
Files.walk API is available from Java 8.
try (Stream<Path> paths = Files.walk(Paths.get("/home/you/Desktop"))) {
paths
.filter(Files::isRegularFile)
.forEach(System.out::println);
}
The example uses try-with-resources pattern recommended in API guide. It ensures that no matter circumstances the stream will be closed.
Solution 2 - Java
File folder = new File("/Users/you/folder/");
File[] listOfFiles = folder.listFiles();
for (File file : listOfFiles) {
if (file.isFile()) {
System.out.println(file.getName());
}
}
Solution 3 - Java
In Java 8 you can do this
Files.walk(Paths.get("/path/to/folder"))
.filter(Files::isRegularFile)
.forEach(System.out::println);
which will print all files in a folder while excluding all directories. If you need a list, the following will do:
Files.walk(Paths.get("/path/to/folder"))
.filter(Files::isRegularFile)
.collect(Collectors.toList())
If you want to return List<File>
instead of List<Path>
just map it:
List<File> filesInFolder = Files.walk(Paths.get("/path/to/folder"))
.filter(Files::isRegularFile)
.map(Path::toFile)
.collect(Collectors.toList());
You also need to make sure to close the stream! Otherwise you might run into an exception telling you that too many files are open. Read here for more information.
Solution 4 - Java
All of the answers on this topic that make use of the new Java 8 functions are neglecting to close the stream. The example in the accepted answer should be:
try (Stream<Path> filePathStream=Files.walk(Paths.get("/home/you/Desktop"))) {
filePathStream.forEach(filePath -> {
if (Files.isRegularFile(filePath)) {
System.out.println(filePath);
}
});
}
From the javadoc of the Files.walk
method:
> 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.
Solution 5 - Java
One remark according to get all files in the directory.
The method Files.walk(path)
will return all files by walking the file tree rooted at the given started file.
For instance, there is the next file tree:
\---folder
| file1.txt
| file2.txt
|
\---subfolder
file3.txt
file4.txt
Using the java.nio.file.Files.walk(Path)
:
Files.walk(Paths.get("folder"))
.filter(Files::isRegularFile)
.forEach(System.out::println);
Gives the following result:
folder\file1.txt
folder\file2.txt
folder\subfolder\file3.txt
folder\subfolder\file4.txt
To get all files only in the current directory use the java.nio.file.Files.list(Path)
:
Files.list(Paths.get("folder"))
.filter(Files::isRegularFile)
.forEach(System.out::println);
Result:
folder\file1.txt
folder\file2.txt
Solution 6 - Java
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());
}
}
}
}
}
Solution 7 - Java
In Java 7 and higher you can use listdir
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);
}
You can also create a filter that can then be passed into the newDirectoryStream
method above
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;
}
}
};
For other filtering examples, [see documentation].(http://docs.oracle.com/javase/tutorial/essential/io/dirs.html#glob)
Solution 8 - Java
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
}
}
Solution 9 - Java
Just walk through all Files using Files.walkFileTree
(Java 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;
}
});
Solution 10 - Java
If you want more options, you can use this function which aims to populate an arraylist of files present in a folder. Options are : recursivility and pattern to match.
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;
}
Best, Adrien
Solution 11 - Java
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() + "\n";
}
}
Solution 12 - Java
nice usage of java.io.FileFilter
as seen on https://stackoverflow.com/a/286001/146745
File fl = new File(dir);
File[] files = fl.listFiles(new FileFilter() {
public boolean accept(File file) {
return file.isFile();
}
});
Solution 13 - Java
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]);
}
}
}
Solution 14 - Java
I think this is good way to read all the files in a folder and sub folder's
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);
}
}
Solution 15 - Java
Simple example that works with Java 1.7 to recursively list files in directories specified on the command-line:
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());
}
}
}
}
Solution 16 - Java
While I do agree with Rich, Orian and the rest for using:
final File keysFileFolder = new File(<path>);
File[] fileslist = keysFileFolder.listFiles();
if(fileslist != null)
{
//Do your thing here...
}
for some reason all the examples here uses absolute path (i.e. all the way from root, or, say, drive letter (C:\) for windows..)
I'd like to add that it is possible to use relative path as-well.
So, if you're pwd (current directory/folder) is folder1 and you want to parse folder1/subfolder, you simply write (in the code above instead of
final File keysFileFolder = new File("subfolder");
Solution 17 - Java
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]);
}
}
}
}
Solution 18 - Java
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);
}
}
}
Solution 19 - Java
Java 8 Files.walk(..)
is good when you are soore it will not throw https://stackoverflow.com/questions/44007055/avoid-java-8-files-walk-termination-cause-of-java-nio-file-accessdeniedexc?noredirect=1&lq=1 .
Here is a safe solution , not though so elegant as Java 8Files.walk(..)
:
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\n", file);
++count[0];
return FileVisitResult.CONTINUE;
}
@Override
public FileVisitResult visitFileFailed(Path file , IOException e) throws IOException {
System.err.printf("Visiting failed for %s\n", file);
return FileVisitResult.SKIP_SUBTREE;
}
@Override
public FileVisitResult preVisitDirectory(Path dir , BasicFileAttributes attrs) throws IOException {
System.out.printf("About to visit directory %s\n", dir);
return FileVisitResult.CONTINUE;
}
});
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Solution 20 - Java
Just to expand on the accepted answer I store the filenames to an ArrayList (instead of just dumping them to System.out.println) I created a helper class "MyFileUtils" so it could be imported by other projects:
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() );
}
}
}
}
I added the full path to the file name. You would use it like this:
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);
}
The ArrayList is passed by "value", but the value is used to point to the same ArrayList object living in the JVM Heap. In this way, each recursion call adds filenames to the same ArrayList (we are NOT creating a new ArrayList on each recursive call).
Solution 21 - Java
There are many good answers above, here's a different approach: In a maven project, everything you put in the resources folder is copied by default in the target/classes folder. To see what is available at runtime
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());
}
Now to get the files from a specific folder, let's say you have a folder called 'res' in your resources folder, just replace:
URL resource = contextClassLoader.getResource("res");
If you want to have access in your com.companyName package then:
contextClassLoader.getResource("com.companyName");
Solution 22 - Java
You can put the file path to argument and create a list with all the filepaths and not put it the list manually. Then use a for loop and a reader. Example for txt files:
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
}}}
Solution 23 - Java
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());
}
}
}
}
Solution 24 - Java
We can use org.apache.commons.io.FileUtils, use listFiles() mehtod to read all the files in a given folder.
eg:
FileUtils.listFiles(directory, new String[] {"ext1", "ext2"}, true)
This read all the files in the given directory with given extensions, we can pass multiple extensions in the array and read recursively within the folder(true parameter).
Solution 25 - Java
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);
}
}
}
}
}
Solution 26 - Java
/**
* 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"));
}
}
You can filter any textfiles or any other extension ..just replace it with .MP3
Solution 27 - Java
> list down files from Test folder present inside class path
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()));
}
}
}
}
}
Solution 28 - Java
This will Read Specified file extension files in given path(looks sub folders also)
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;
}
Solution 29 - Java
This will work fine:
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);
}
}
Solution 30 - Java
Given a baseDir, lists out all the files and directories below it, written iteratively.
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;
}
Solution 31 - Java
public static List<File> files(String dirname) {
if (dirname == null) {
return Collections.emptyList();
}
File dir = new File(dirname);
if (!dir.exists()) {
return Collections.emptyList();
}
if (!dir.isDirectory()) {
return Collections.singletonList(file(dirname));
}
return Arrays.stream(Objects.requireNonNull(dir.listFiles()))
.collect(Collectors.toList());
}
Solution 32 - Java
to prevent Nullpointerexceptions on the listFiles() function and recursivly get all files from subdirectories too..
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);
Solution 33 - Java
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());
}
}
}
}
}
Just pass your folder it will tell you main class about the method.