How can I get list all the files within a folder recursively in Java?
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what do you mean by TREE? please rephrase your question if you're seeking a method to LIST all files within a folder (recursively).Peter Perháč– Peter Perháč2012-05-08 10:10:21 +00:00Commented May 8, 2012 at 10:10
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@PeterPerháč this question is 4 years old.. :) thanksLipis– Lipis2012-05-08 11:58:31 +00:00Commented May 8, 2012 at 11:58
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6 Answers
Not sure how you want to represent the tree? Anyway here's an example which scans the entire subtree using recursion. Files and directories are treated alike. Note that File.listFiles() returns null for non-directories.
public static void main(String[] args) {
Collection<File> all = new ArrayList<File>();
addTree(new File("."), all);
System.out.println(all);
}
static void addTree(File file, Collection<File> all) {
File[] children = file.listFiles();
if (children != null) {
for (File child : children) {
all.add(child);
addTree(child, all);
}
}
}
Java 7 offers a couple of improvements. For example, DirectoryStream provides one result at a time - the caller no longer has to wait for all I/O operations to complete before acting. This allows incremental GUI updates, early cancellation, etc.
static void addTree(Path directory, Collection<Path> all)
throws IOException {
try (DirectoryStream<Path> ds = Files.newDirectoryStream(directory)) {
for (Path child : ds) {
all.add(child);
if (Files.isDirectory(child)) {
addTree(child, all);
}
}
}
}
Note that the dreaded null return value has been replaced by IOException.
Java 7 also offers a tree walker:
static void addTree(Path directory, final Collection<Path> all)
throws IOException {
Files.walkFileTree(directory, new SimpleFileVisitor<Path>() {
@Override
public FileVisitResult visitFile(Path file, BasicFileAttributes attrs)
throws IOException {
all.add(file);
return FileVisitResult.CONTINUE;
}
});
}
import java.io.File;
public class Test {
public static void main( String [] args ) {
File actual = new File(".");
for( File f : actual.listFiles()){
System.out.println( f.getName() );
}
}
}
It displays indistinctly files and folders.
See the methods in File class to order them or avoid directory print etc.
Comments
You can also use the FileFilter interface to filter out what you want. It is best used when you create an anonymous class that implements it:
import java.io.File;
import java.io.FileFilter;
public class ListFiles {
public File[] findDirectories(File root) {
return root.listFiles(new FileFilter() {
public boolean accept(File f) {
return f.isDirectory();
}});
}
public File[] findFiles(File root) {
return root.listFiles(new FileFilter() {
public boolean accept(File f) {
return f.isFile();
}});
}
}
Comments
public static void directory(File dir) {
File[] files = dir.listFiles();
for (File file : files) {
System.out.println(file.getAbsolutePath());
if (file.listFiles() != null)
directory(file);
}
}
Here dir is Directory to be scanned. e.g. c:\
Comments
Visualizing the tree structure was the most convenient way for me :
public static void main(String[] args) throws IOException {
printTree(0, new File("START/FROM/DIR"));
}
static void printTree(int depth, File file) throws IOException {
StringBuilder indent = new StringBuilder();
String name = file.getName();
for (int i = 0; i < depth; i++) {
indent.append(".");
}
//Pretty print for directories
if (file.isDirectory()) {
System.out.println(indent.toString() + "|");
if(isPrintName(name)){
System.out.println(indent.toString() + "*" + file.getName() + "*");
}
}
//Print file name
else if(isPrintName(name)) {
System.out.println(indent.toString() + file.getName());
}
//Recurse children
if (file.isDirectory()) {
File[] files = file.listFiles();
for (int i = 0; i < files.length; i++){
printTree(depth + 4, files[i]);
}
}
}
//Exclude some file names
static boolean isPrintName(String name){
if (name.charAt(0) == '.') {
return false;
}
if (name.contains("svn")) {
return false;
}
//.
//. Some more exclusions
//.
return true;
}
Comments
In JDK7, "more NIO features" should have methods to apply the visitor pattern over a file tree or just the immediate contents of a directory - no need to find all the files in a potentially huge directory before iterating over them.
answered Oct 10, 2008 at 8:13
Tom Hawtin - tackline
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2 Comments
Pacerier
but Is NIO faster?
Tom Hawtin - tackline
@Pacerier For small directories it shouldn't make much difference. I'm even less familiar with the implementation now than I was nine years ago, but streaming NIO should be better at handling massive directories than reading all at once. Having said that, memory will now be several dozen times cheaper.