I am trying to filter hidden files using the NIO classes.
When I run the attached code on Windows 10 I get the following output:
Files:
c:\Documents and Settings
c:\PerfLogs
c:\Program Files
c:\Program Files (x86)
c:\Users
c:\Windows
Paths:
c:\$Recycle.Bin
c:\Config.Msi
c:\Documents and Settings
c:\Intel
c:\IntelOptaneData
c:\OEM
c:\OneDriveTemp
c:\PerfLogs
c:\Program Files
c:\Program Files (x86)
c:\ProgramData
c:\Recovery
c:\System Volume Information
c:\Users
c:\Windows
The list displayed under Files (using the old File.listFiles(FileFilter) method) is the list I see in Windows File Explorer and is what I am expecting to see (except for the Document and Setting and I know how to fix that)
Why is the NIO approach not filtering hidden files the same way?
How do I get NIO filtering to be the same?
Here is the test code:
import java.io.*;
import java.nio.file.*;
public class ListFilesNIO
{
public static void main(String[] args) throws Exception
{
String directory = "c:\\";
// Use old File I/O
FileFilter fileFilter = new FileFilter()
{
#Override
public boolean accept(File entry)
{
if (entry.isHidden()) return false;
return true;
}
};
System.out.println("Files:");
File[] files = new File( directory ).listFiles( fileFilter );
for (File file : files)
{
System.out.println( "\t" + file );
}
// Use NIO
DirectoryStream.Filter<Path> pathFilter = new DirectoryStream.Filter<Path>()
{
#Override
public boolean accept(Path entry) throws IOException
{
if (Files.isHidden( entry )) return false;
return true;
}
};
System.out.println();
System.out.println("Paths:");
DirectoryStream<Path> paths = Files.newDirectoryStream(Paths.get( directory ), pathFilter);
for (Path path : paths)
{
System.out.println( "\t" + path );
}
}
}
Note: when I run the code without the filter, in both cases 18 files are displayed. So the first approach is filtering 12 hidden files and the second approach is only filtering 3 files.
It's not a bug but a feature(!) known since jdk7, Windows hidden directory are not detected as hidden, see this bug and this one (fix jdk13).
As a workaround, you can do this :
import java.nio.file.attribute.DosFileAttributes;
...
DirectoryStream.Filter<Path> pathFilter = new DirectoryStream.Filter<Path>()
{
#Override
public boolean accept(Path entry) throws IOException
{
DosFileAttributes attr = Files.readAttributes(entry, DosFileAttributes.class);
return !attr.isHidden();
}
};
I ended up using:
DirectoryStream.Filter<Path> pathFilter = new DirectoryStream.Filter<Path>()
{
#Override
public boolean accept(Path entry) throws IOException
{
DosFileAttributes attr = Files.readAttributes(entry, DosFileAttributes.class, LinkOption.NOFOLLOW_LINKS);
return !attr.isHidden();
}
};
As I mentioned in my question, I also want the Documents and Settings to be hidden.
The Documents and Settings is a link to C:\Users.
The default implementation for the Files.readAttributes(…) method is to follow links. So I guess because the c:\Users directory is not hidden, the Documents and Settings is also considered not hidden.
By using LinkOption.NOFOLLOW_LINKS it is considered hidden, which is what I want.
Whats wrong with my code ? with reference with this question
the folder is not being deleted.
File f = new File(directory+"\\OOO");
if (f.exists())
{
System.out.println(f);
FileUtils.delete(f);
f.delete();
// f.mkdir();
}
else
{
System.out.println("created");
//f.mkdir();
}
FileUtils does not contain any method named delete();
You should use deleteDirectory() to recursively delete directory.
Or you can also use deleteQuitely() to suppress any exceptions while deletion.
FileUtils does not contain any method named delete(), so you should use deleteDirectory() instead. Also, please ensure you have imported the correct Java file utility class from Apache Commons IO. If it is not the case, your need to
Download Apache Commons IO as a JAR file, link it to the classpath
Or declare a dependency the Apache Commons IO in your Maven, or other dependency management system.
Here's the code:
import org.apache.commons.io.FileUtils;
public class YourClass
{
public void yourMethod() throws IOException
{
File f = new File(directory+"\\OOO");
if (f.exists())
{
FileUtils.deleteDirectory(f);
}
...
}
}
This may help
try {
FileUtils.deleteDirectory(new File("path");
//path Example - C:\\myfolder\\foldertodelete
} catch (IOException e) {
}
I am looking for a way to get a list of all resource names from a given classpath directory, something like a method List<String> getResourceNames (String directoryName).
For example, given a classpath directory x/y/z containing files a.html, b.html, c.html and a subdirectory d, getResourceNames("x/y/z") should return a List<String> containing the following strings:['a.html', 'b.html', 'c.html', 'd'].
It should work both for resources in filesystem and jars.
I know that I can write a quick snippet with Files, JarFiles and URLs, but I do not want to reinvent the wheel. My question is, given existing publicly available libraries, what is the quickest way to implement getResourceNames? Spring and Apache Commons stacks are both feasible.
Custom Scanner
Implement your own scanner. For example:
(limitations of this solution are mentioned in the comments)
private List<String> getResourceFiles(String path) throws IOException {
List<String> filenames = new ArrayList<>();
try (
InputStream in = getResourceAsStream(path);
BufferedReader br = new BufferedReader(new InputStreamReader(in))) {
String resource;
while ((resource = br.readLine()) != null) {
filenames.add(resource);
}
}
return filenames;
}
private InputStream getResourceAsStream(String resource) {
final InputStream in
= getContextClassLoader().getResourceAsStream(resource);
return in == null ? getClass().getResourceAsStream(resource) : in;
}
private ClassLoader getContextClassLoader() {
return Thread.currentThread().getContextClassLoader();
}
Spring Framework
Use PathMatchingResourcePatternResolver from Spring Framework.
Ronmamo Reflections
The other techniques might be slow at runtime for huge CLASSPATH values. A faster solution is to use ronmamo's Reflections API, which precompiles the search at compile time.
Here is the code
Source: forums.devx.com/showthread.php?t=153784
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Enumeration;
import java.util.regex.Pattern;
import java.util.zip.ZipEntry;
import java.util.zip.ZipException;
import java.util.zip.ZipFile;
/**
* list resources available from the classpath # *
*/
public class ResourceList{
/**
* for all elements of java.class.path get a Collection of resources Pattern
* pattern = Pattern.compile(".*"); gets all resources
*
* #param pattern
* the pattern to match
* #return the resources in the order they are found
*/
public static Collection<String> getResources(
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final String classPath = System.getProperty("java.class.path", ".");
final String[] classPathElements = classPath.split(System.getProperty("path.separator"));
for(final String element : classPathElements){
retval.addAll(getResources(element, pattern));
}
return retval;
}
private static Collection<String> getResources(
final String element,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final File file = new File(element);
if(file.isDirectory()){
retval.addAll(getResourcesFromDirectory(file, pattern));
} else{
retval.addAll(getResourcesFromJarFile(file, pattern));
}
return retval;
}
private static Collection<String> getResourcesFromJarFile(
final File file,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
ZipFile zf;
try{
zf = new ZipFile(file);
} catch(final ZipException e){
throw new Error(e);
} catch(final IOException e){
throw new Error(e);
}
final Enumeration e = zf.entries();
while(e.hasMoreElements()){
final ZipEntry ze = (ZipEntry) e.nextElement();
final String fileName = ze.getName();
final boolean accept = pattern.matcher(fileName).matches();
if(accept){
retval.add(fileName);
}
}
try{
zf.close();
} catch(final IOException e1){
throw new Error(e1);
}
return retval;
}
private static Collection<String> getResourcesFromDirectory(
final File directory,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final File[] fileList = directory.listFiles();
for(final File file : fileList){
if(file.isDirectory()){
retval.addAll(getResourcesFromDirectory(file, pattern));
} else{
try{
final String fileName = file.getCanonicalPath();
final boolean accept = pattern.matcher(fileName).matches();
if(accept){
retval.add(fileName);
}
} catch(final IOException e){
throw new Error(e);
}
}
}
return retval;
}
/**
* list the resources that match args[0]
*
* #param args
* args[0] is the pattern to match, or list all resources if
* there are no args
*/
public static void main(final String[] args){
Pattern pattern;
if(args.length < 1){
pattern = Pattern.compile(".*");
} else{
pattern = Pattern.compile(args[0]);
}
final Collection<String> list = ResourceList.getResources(pattern);
for(final String name : list){
System.out.println(name);
}
}
}
If you are using Spring Have a look at PathMatchingResourcePatternResolver
Using Reflections
Get everything on the classpath:
Reflections reflections = new Reflections(null, new ResourcesScanner());
Set<String> resourceList = reflections.getResources(x -> true);
Another example - get all files with extension .csv from some.package:
Reflections reflections = new Reflections("some.package", new ResourcesScanner());
Set<String> resourceList = reflections.getResources(Pattern.compile(".*\\.csv"));
So in terms of the PathMatchingResourcePatternResolver this is what is needed in the code:
#Autowired
ResourcePatternResolver resourceResolver;
public void getResources() {
resourceResolver.getResources("classpath:config/*.xml");
}
If you use apache commonsIO you can use for the filesystem (optionally with extension filter):
Collection<File> files = FileUtils.listFiles(new File("directory/"), null, false);
and for resources/classpath:
List<String> files = IOUtils.readLines(MyClass.class.getClassLoader().getResourceAsStream("directory/"), Charsets.UTF_8);
If you don't know if "directoy/" is in the filesystem or in resources you may add a
if (new File("directory/").isDirectory())
or
if (MyClass.class.getClassLoader().getResource("directory/") != null)
before the calls and use both in combination...
The most robust mechanism for listing all resources in the classpath is currently to use this pattern with ClassGraph, because it handles the widest possible array of classpath specification mechanisms, including the new JPMS module system. (I am the author of ClassGraph.)
List<String> resourceNames;
try (ScanResult scanResult = new ClassGraph().acceptPaths("x/y/z").scan()) {
resourceNames = scanResult.getAllResources().getNames();
}
The Spring framework's PathMatchingResourcePatternResolver is really awesome for these things:
private Resource[] getXMLResources() throws IOException
{
ClassLoader classLoader = MethodHandles.lookup().getClass().getClassLoader();
PathMatchingResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(classLoader);
return resolver.getResources("classpath:x/y/z/*.xml");
}
Maven dependency:
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-core</artifactId>
<version>LATEST</version>
</dependency>
This should work (if spring is not an option):
public static List<String> getFilenamesForDirnameFromCP(String directoryName) throws URISyntaxException, UnsupportedEncodingException, IOException {
List<String> filenames = new ArrayList<>();
URL url = Thread.currentThread().getContextClassLoader().getResource(directoryName);
if (url != null) {
if (url.getProtocol().equals("file")) {
File file = Paths.get(url.toURI()).toFile();
if (file != null) {
File[] files = file.listFiles();
if (files != null) {
for (File filename : files) {
filenames.add(filename.toString());
}
}
}
} else if (url.getProtocol().equals("jar")) {
String dirname = directoryName + "/";
String path = url.getPath();
String jarPath = path.substring(5, path.indexOf("!"));
try (JarFile jar = new JarFile(URLDecoder.decode(jarPath, StandardCharsets.UTF_8.name()))) {
Enumeration<JarEntry> entries = jar.entries();
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
String name = entry.getName();
if (name.startsWith(dirname) && !dirname.equals(name)) {
URL resource = Thread.currentThread().getContextClassLoader().getResource(name);
filenames.add(resource.toString());
}
}
}
}
}
return filenames;
}
My way, no Spring, used during a unit test:
URI uri = TestClass.class.getResource("/resources").toURI();
Path myPath = Paths.get(uri);
Stream<Path> walk = Files.walk(myPath, 1);
for (Iterator<Path> it = walk.iterator(); it.hasNext(); ) {
Path filename = it.next();
System.out.println(filename);
}
With Spring it's easy. Be it a file, or folder, or even multiple files, there are chances, you can do it via injection.
This example demonstrates the injection of multiple files located in x/y/z folder.
import org.springframework.beans.factory.annotation.Value;
import org.springframework.core.io.Resource;
import org.springframework.stereotype.Service;
#Service
public class StackoverflowService {
#Value("classpath:x/y/z/*")
private Resource[] resources;
public List<String> getResourceNames() {
return Arrays.stream(resources)
.map(Resource::getFilename)
.collect(Collectors.toList());
}
}
It does work for resources in the filesystem as well as in JARs.
Used a combination of Rob's response.
final String resourceDir = "resourceDirectory/";
List<String> files = IOUtils.readLines(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir), Charsets.UTF_8);
for (String f : files) {
String data = IOUtils.toString(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir + f));
// ... process data
}
I think you can leverage the [Zip File System Provider][1] to achieve this. When using FileSystems.newFileSystem it looks like you can treat the objects in that ZIP as a "regular" file.
In the linked documentation above:
Specify the configuration options for the zip file system in the java.util.Map object passed to the FileSystems.newFileSystem method. See the [Zip File System Properties][2] topic for information about the provider-specific configuration properties for the zip file system.
Once you have an instance of a zip file system, you can invoke the methods of the [java.nio.file.FileSystem][3] and [java.nio.file.Path][4] classes to perform operations such as copying, moving, and renaming files, as well as modifying file attributes.
The documentation for the jdk.zipfs module in [Java 11 states][5]:
The zip file system provider treats a zip or JAR file as a file system and provides the ability to manipulate the contents of the file. The zip file system provider can be created by [FileSystems.newFileSystem][6] if installed.
Here is a contrived example I did using your example resources. Note that a .zip is a .jar, but you could adapt your code to instead use classpath resources:
Setup
cd /tmp
mkdir -p x/y/z
touch x/y/z/{a,b,c}.html
echo 'hello world' > x/y/z/d
zip -r example.zip x
Java
import java.io.IOException;
import java.net.URI;
import java.nio.file.FileSystem;
import java.nio.file.FileSystems;
import java.nio.file.Files;
import java.util.Collections;
import java.util.stream.Collectors;
public class MkobitZipRead {
public static void main(String[] args) throws IOException {
final URI uri = URI.create("jar:file:/tmp/example.zip");
try (
final FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.emptyMap());
) {
Files.walk(zipfs.getPath("/")).forEach(path -> System.out.println("Files in zip:" + path));
System.out.println("-----");
final String manifest = Files.readAllLines(
zipfs.getPath("x", "y", "z").resolve("d")
).stream().collect(Collectors.joining(System.lineSeparator()));
System.out.println(manifest);
}
}
}
Output
Files in zip:/
Files in zip:/x/
Files in zip:/x/y/
Files in zip:/x/y/z/
Files in zip:/x/y/z/c.html
Files in zip:/x/y/z/b.html
Files in zip:/x/y/z/a.html
Files in zip:/x/y/z/d
-----
hello world
Neither of answers worked for me even though I had my resources put in resources folders and followed the above answers. What did make a trick was:
#Value("file:*/**/resources/**/schema/*.json")
private Resource[] resources;
Expanding on Luke Hutchinsons answer above, using his ClassGraph library, I was able to easily get a list of all files in a Resource folder with almost no effort at all.
Let's say that in your resource folder, you have a folder called MyImages. This is how easy it is to get a URL list of all the files in that folder:
import io.github.classgraph.ClassGraph;
import io.github.classgraph.ResourceList;
import io.github.classgraph.ScanResult;
public static LinkedList<URL> getURLList (String folder) {
LinkedList<URL> urlList = new LinkedList<>();
ScanResult scanResult = new ClassGraph().enableAllInfo().scan();
ResourceList resources = scanResult.getAllResources();
for (URL url : resources.getURLs()) {
if (url.toString().contains(folder)) {
urlList.addLast(url);
}
}
return urlList;
}
Then you simply do this:
LinkedList<URL> myURLFileList = getURLList("MyImages");
The URLs can then be loaded into streams or use Apache's FileUtils to copy the files somewhere else like this:
String outPath = "/My/Output/Path";
for(URL url : myURLFileList) {
FileUtils.copyURLToFile(url, new File(outPath, url.getFile()));
}
I think ClassGraph is a pretty slick library for making tasks like this very simple and easy to comprehend.
Based on #rob 's information above, I created the implementation which I am releasing to the public domain:
private static List<String> getClasspathEntriesByPath(String path) throws IOException {
InputStream is = Main.class.getClassLoader().getResourceAsStream(path);
StringBuilder sb = new StringBuilder();
while (is.available()>0) {
byte[] buffer = new byte[1024];
sb.append(new String(buffer, Charset.defaultCharset()));
}
return Arrays
.asList(sb.toString().split("\n")) // Convert StringBuilder to individual lines
.stream() // Stream the list
.filter(line -> line.trim().length()>0) // Filter out empty lines
.collect(Collectors.toList()); // Collect remaining lines into a List again
}
While I would not have expected getResourcesAsStream to work like that on a directory, it really does and it works well.
I am playing a bit with the new Java 7 IO features. Actually I am trying to retrieve all the XML files in a folder. However this throws an exception when the folder does not exist. How can I check if the folder exists using the new IO?
public UpdateHandler(String release) {
log.info("searching for configuration files in folder " + release);
Path releaseFolder = Paths.get(release);
try(DirectoryStream<Path> stream = Files.newDirectoryStream(releaseFolder, "*.xml")){
for (Path entry: stream){
log.info("working on file " + entry.getFileName());
}
}
catch (IOException e){
log.error("error while retrieving update configuration files " + e.getMessage());
}
}
Using java.nio.file.Files:
Path path = ...;
if (Files.exists(path)) {
// ...
}
You can optionally pass this method LinkOption values:
if (Files.exists(path, LinkOption.NOFOLLOW_LINKS)) {
There's also a method notExists:
if (Files.notExists(path)) {
Quite simple:
new File("/Path/To/File/or/Directory").exists();
And if you want to be certain it is a directory:
File f = new File("/Path/To/File/or/Directory");
if (f.exists() && f.isDirectory()) {
...
}
To check if a directory exists with the new IO:
if (Files.isDirectory(Paths.get("directory"))) {
...
}
isDirectory returns true if the file is a directory; false if the file does not exist, is not a directory, or it cannot be determined if the file is a directory or not.
See: documentation.
Generate a file from the string of your folder directory
String path="Folder directory";
File file = new File(path);
and use method exist.
If you want to generate the folder you sould use mkdir()
if (!file.exists()) {
System.out.print("No Folder");
file.mkdir();
System.out.print("Folder created");
}
You need to transform your Path into a File and test for existence:
for(Path entry: stream){
if(entry.toFile().exists()){
log.info("working on file " + entry.getFileName());
}
}
There is no need to separately call the exists() method, as isDirectory() implicitly checks whether the directory exists or not.
import java.io.File;
import java.nio.file.Paths;
public class Test
{
public static void main(String[] args)
{
File file = new File("C:\\Temp");
System.out.println("File Folder Exist" + isFileDirectoryExists(file));
System.out.println("Directory Exists" + isDirectoryExists("C:\\Temp"));
}
public static boolean isFileDirectoryExists(File file)
{
if (file.exists())
{
return true;
}
return false;
}
public static boolean isDirectoryExists(String directoryPath)
{
if (!Paths.get(directoryPath).toFile().isDirectory())
{
return false;
}
return true;
}
}
We can check files and thire Folders.
import java.io.*;
public class fileCheck
{
public static void main(String arg[])
{
File f = new File("C:/AMD");
if (f.exists() && f.isDirectory()) {
System.out.println("Exists");
//if the file is present then it will show the msg
}
else{
System.out.println("NOT Exists");
//if the file is Not present then it will show the msg
}
}
}
File sourceLoc=new File("/a/b/c/folderName");
boolean isFolderExisted=false;
sourceLoc.exists()==true?sourceLoc.isDirectory()==true?isFolderExisted=true:isFolderExisted=false:isFolderExisted=false;
From SonarLint, if you already have the path, use path.toFile().exists() instead of Files.exists for better performance.
The Files.exists method has noticeably poor performance in JDK 8, and can slow an application significantly when used to check files that don't actually exist.
The same goes for Files.notExists, Files.isDirectory and Files.isRegularFile.
Noncompliant Code Example:
Path myPath;
if(java.nio.Files.exists(myPath)) { // Noncompliant
// do something
}
Compliant Solution:
Path myPath;
if(myPath.toFile().exists())) {
// do something
}
Here is the code I have thus far:
import java.io.*;
class JAVAFilter implements FilenameFilter {
public boolean accept(File dir, String name) {
return (name.endsWith(".java"));
}
}
public class tester {
public static void main(String args[])
{
FilenameFilter filter = new JAVAFilter();
File directory = new File("C:\\1.3\\");
String filename[] = directory.list(filter);
}
}
At this point, it'll store a list of all the *.java files from the directory C:\1.3\ in the string array filename. However, i'd like to store a list of all the java files also in subdirectories (preferably with their path within C:\1.3\ specified also. How do I go about doing this? Thanks!
you should look at DirectoryWalker from Apache
I'm afraid you can't do it with the list(FilenameFilter) method. You'll have to list all files and directories, and then do the filtering yourself. Something like this:
public List<File> getFiles(File dir, FilenameFilter filter) {
List<File> ret = new ArrayList<File>();
for (File f : dir.listFiles()) {
if (f.isDirectory()) {
ret.addAll(getFiles(f, filter));
} else if (filter.accept(dir, f.getName())) {
ret.add(f);
}
}
return ret;
}
As far as I know, you will have to do this manually (recursively), i.e. you will have to call list(filter) for all sub-directories of C:\1.3\, and so on....