public void saveFile(MultipartFile file , int id) throws IOException {
String date = LocalDateTime.now().format(DateTimeFormatter.ofPattern("-yyMMddHHmmss"));
String destination = goldenFileLocation + "/" + file.getOriginalFilename() + date;
file.transferTo(new File(destination));
}
I need to write a Mockito test for the above service layer code. I have the below code but not sure how to test if the file exists in the mentioned location.
#Test
public void testSaveFile() throws IOException {
MockMultipartFile firstFile = new MockMultipartFile("pdf", "response.pdf", "multipart/form-data", "data".getBytes());
saveFile(firstFile,1);
//check if file got stored in the mentioned location
}
I found two ways to create temporary directories in JUnit.
Way 1:
#Rule
public TemporaryFolder tempDirectory = new TemporaryFolder();
#Test
public void testTempDirectory() throws Exception {
tempDirectory.newFile("test.txt");
tempDirectory.newFolder("myDirectory");
// how do I add files to myDirectory?
}
Way 2:
#Test
public void testTempDirectory() throws Exception {
File myFile = File.createTempFile("abc", "txt");
File myDirectory = Files.createTempDir();
// how do I add files to myDirectory?
}
As the comment above mentions, I have a requirement where I want to add some temporary files in these temporary directories. Run my test against this structure and finally delete everything on exit.
How do I do that?
You can do it the same way you do it for real folders.
#Rule
public TemporaryFolder rootFolder = new TemporaryFolder();
#Test
public void shouldCreateChildFile() throws Exception {
File myFolder = rootFolder.newFolder("my-folder");
File myFile = new File(myFolder, "my-file.txt");
}
Using new File(subFolderOfTemporaryFolder, "fileName") did not work for me. Calling subFolder.list() returned an empty array. This is how I made it work:
#Rule
public TemporaryFolder temporaryFolder = new TemporaryFolder();
#Test
public void createFileInSubFolderOfTemporaryFolder() throws IOException {
String subFolderName = "subFolder";
File subFolder = temporaryFolder.newFolder(subFolderName);
temporaryFolder.newFile(subFolderName + File.separator + "fileName1");
String[] actual = subFolder.list();
assertFalse(actual.length == 0);
}
Using TemporaryFolder creates a directory with a common root. Once you have created a folder, you can then create a file by specifying the directory structure and the final filename as the name.
#Rule
public TemporaryFolder rootFolder = new TemporaryFolder();
#Test
public void shouldCreateChildFile() throws Exception {
File myFolder = rootFolder.newFolder("my-folder");
File myFileInMyFolder = rootFolder.newFile("/my-folder/my-file.txt");
}
You can create child directories in the same way.
There two ways to delete temp directory or temp file.Fist,delete the dirctory or file manually use file.delete() method,Second,delete the temp directory or file when program exis user file.deleteOnExist().
You can try this,I print the path to console,you can to check realy delte or not,I test on windows7 system.
File myDirectory = Files.createTempDir();
File tmpFile = new File(myDirectory.getAbsolutePath() + File.separator + "test.txt");
FileUtils.writeStringToFile(tmpFile, "HelloWorld", "UTF-8");
System.out.println(myDirectory.getAbsolutePath());
// clean
tmpFile.delete();
myDirectory.deleteOnExit();
Hi guys I'm trying to find if a specific file is inside the project directory.
File f = new File(System.getProperty("user.dir"));
System.out.println(f);
String archivoLiga="LigaV2";
System.out.println(f.listFiles((dir1, name) -> name.startsWith(archivoLiga) && name.endsWith(".properties")).length == 0);
But this only works if the file is in the "first" level, i want it to find it even if it's inside another folder. Any ideas?
Use Java 8's find() method to recurse subdirectories:
final int MAX_DEPTH = 50; // Max depth of subdirectories to search
Path userDir = Paths.get(System.getProperty("user.dir"));
System.out.println(userDir);
String archivoLiga="LigaV2";
System.out.println(
Files.find(
userDir,
maxDepth,
(path,attr) -> path.getFileName().startsWith(archivoLiga)
&& path.getFileName().endsWith(".properties"))
.findAny()
.isPresent());
Try recursively looking inside subfolders :-
public boolean checkForFile(String dirname,String prefix,String ext){
File dir = new File(dirname);
//System.out.println(dir);
for(File f : dir.listFiles()){
if(f.isFile()){
if(f.getName().startsWith(prefix) && f.getName().endsWith(ext)){
System.out.println(f.getName());
return true;
}
}
else{
//This step starts looking inside subfolder as well
return checkForFile(f.getAbsolutePath(),prefix,ext);
}
}
return true;
}
To check a file inside a folder, You will need to use exists() method of java.io.File like this:
boolean exists = new File("FOLDER_PATH/FILE_NAME").exists();
if (exists) {
System.out.println("File exists inside given folder");
} else {
System.out.println("File does not exists inside given folder");
}
Also possible with FileVisitor:
#Getter #Setter
#RequiredArgsConstructor
public static class SearchVisitor extends SimpleFileVisitor<Path> {
private final String fileToSearch;
private boolean found=false;
#Override
public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) throws IOException {
if(!file.getFileName().toString().equals(fileToSearch)) return FileVisitResult.CONTINUE;
found=true;
return FileVisitResult.TERMINATE;
}
}
public void test() throws IOException {
SearchVisitor sv = new SearchVisitor("LigaV2");
Files.walkFileTree( Paths.get(System.getProperty("user.dir")), sv);
log.info("found file {}:{}", sv.getFileToSearch(), sv.isFound());
}
Use public boolean isDirectory() in java file API following is the link to oracle documentation. Tests whether the file denoted by this abstract pathname is a directory.
https://docs.oracle.com/javase/7/docs/api/java/io/File.html#isDirectory()
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 did this simple experiment to list all files/directory in a parent directory.
Did this by making a java project in eclipse by name 'JavaProject' and a class 'Temp.java' under src/com. Code is as below:
public class Temp {
public static void main(String[] args) {
search("../JavaProject");
}
public static void search(String dName) {
String[] files = new String[100];
File search = new File(dName); // make file object
if (!search.isDirectory()) {
return;
}
files = search.list(); // create the list
for (String fn : files) {// iterate through it
System.out.print(" " + fn);
File temp = new File(fn);
if (temp.isDirectory()) {
search(fn);
}
System.out.println();
}
}
}
The file structure is as below :
JavaProject(dir)
.classpath(file)
.project(file)
.settings(dir)
org.eclipse.jdt.core.prefs(file)
bin(dir)
com(file)
Temp.class(file)
src(dir)
com(dir)
Temp.java(file)
When I run the above program, it gives the following output:
.classpath
.project
.settings org.eclipse.jdt.core.prefs
bin com
src com
I cant understand why it does not print the .java file and .class file inside the com folders.
When I try debugging then the file object on 'com' returns 'false' for both isDirectory() and isFile() methods.
When it gets to the 'com' directory your code is doing:
File temp = new File("com");
Since you have not specified any path this will be taken to be relative to the current directory which is not the directory containing 'com'.
You should use something like:
File temp = new File(parent, fn);
where parent is the File object for the parent directory.
You can use listFiles() instead of list(). See below example:
public class Program {
public static void main(String args[]) throws IOException {
search(new File("."), 0);
}
public static void search(File file, int level) {
if (!file.isDirectory()) {
return;
}
for (File f : file.listFiles()) {
for (int i = 0; i < level; i++) {
System.out.print(" ");
}
System.out.println(f.getName());
if (f.isDirectory()) {
search(f, ++level);
}
}
}
}