I am accessing a File inside the resources folder from the main class
File file = new ClassPathResource("remoteUnitsIdsInOldServer.txt").getFile();
and I am getting this error:
java.io.FileNotFoundException: class path resource [remoteUnitsIdsInOldServer.txt] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/Users/lopes/Documents/workspace-sts-3.9.0.RELEASE/telefonicaUtils/target/telefonicaUtils-0.0.1-SNAPSHOT.jar!/BOOT-INF/classes!/remoteUnitsIdsInOldServer.txt
and I even open the jar file and the file remoteUnitsIdsInOldServer.txt is there, inside classes
The simplest solution for me was,
try {
ClassPathResource classPathResource = new ClassPathResource("remoteUnitsIdsInOldServer.txt");
byte[] data = FileCopyUtils.copyToByteArray(classPathResource.getInputStream());
String content = new String(data, StandardCharsets.UTF_8);
} catch (Exception ex) {
ex.printStackTrace();
}
It's depends on your requirements..
Case 1:
Considering you need to access text file from resource. You can simply use apache IOUtils and java ClassLoader.
Snippet (note: IOUtils package --> org.apache.commons.io.IOUtils)
String result = "";
ClassLoader classLoader = getClass().getClassLoader();
try {
result = IOUtils.toString(classLoader.getResourceAsStream("fileName"));
} catch (IOException e) {
e.printStackTrace();
}
Classic way:
StringBuilder result = new StringBuilder("");
//Get file from resources folder
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
try (Scanner scanner = new Scanner(file)) {
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
result.append(line).append("\n");
}
scanner.close();
} catch (IOException e) {
e.printStackTrace();
}
Case 2:
Considering you need to access properties from resources such as xml, properties files.
Its too simple, Simply use spring annotation #ImportResource({ "classpath:application-properties.xml", "classpath:context.properties" })
Hope that will be helpful to you.
Typically, a source tree would look like this:
src/
main/
java/
com/
...
resources/
remoteUnitsIdsInOldServer.txt
And, using standard Maven/Gradle functionality, would produce a JAR like this:
<JAR_ROOT>/
com/
...
remoteUnitsIdsInOldServer.txt
Your listed code should work for this situation. However, you mentioned looking in your JAR "inside classes". I wouldn't think there would be a "classes" folder within the JAR.
Good luck.
In my SpringBoot project I would like to read an image file located in:
I tried:
URL url = getClass().getResource("android.png");
File f = new File(url.getPath());
and:
URL url = getClass().getResource("static/images/android.png");
File f = new File(url.getPath());
but the result is null.
EDIT: this worked
try {
File file = new ClassPathResource("static/images/android.png").getFile();
} catch (IOException e) {
}
If you are using Spring, then it's better to use the built-in ClassPathResource class to access files within your classpath.
You can try this
File file = new ClassPathResource("/images/android.png").getFile();
because we can access static folder from anywhere
you can use this.getClass().getResource("/static/images").getFile(); to get the folder location and then append with the file name to get the file location.
Example
String path=this.getClass().getResource("/static/images").getFile();
File File f = new File(path+"/​android.png");
Have you tried
ClassLoader.getSystemResourceAsStream("/static/images/android.png");
When I getting something in my code I use a short path like
this.getStylesheets().add("/css/editorTool.css");
but when I writing a new file I give exact paht - like:
File f = new File("D:\\IdeaProjects\\SmartCRM for TB\\EditorTool\\resourses\\html\\test.html");
try{
BufferedWriter bw = new BufferedWriter(new FileWriter(f));
bw.write(html);
bw.close();
} catch (IOException e) {
e.printStackTrace();
}
As i will have to distribute this app among some users I will not be able to garanty that the absolute path to my app will be the same. How i can write it to so that it is like "path_to_the_folder_that_is_contained_in_app + fileName?
you can use system property "user.dir" to get current working path, the code is alike this:
String currentWorkingPath = System.getProperty("user.dir");
String fullPath = currentWorkingPath + File.separator+ fileName;
then the file will be under currentWorkingPath when the program run on any PC.
You can use getResource() method of java.lang.Class<T>
The java.lang.Class.getResource() finds a resource with a given name
Documentation Here
usage:
File file = new File(getClass().getResource("html/test.html").getPath());
I created a desktop project in netbeans, in the project folder I have three files : file.txt, file2.txt and file3.txt, in the load of the program I want to call these three files, and this is the code I tried :
public void run() {
Path path = Paths.get("file.txt");
Path path2 = Paths.get("file2.txt");
Path path3 = Paths.get("file3.txt");
if(Files.exists(path) && Files.exists(path2) && Files.exists(path3)) {
lireFichiers();
}else{
JOptionPane.showConfirmDialog(null, "Files didn't found !");
}
}
but when I run my program I get the message : "Files didn't found !" which means he didn't found those files.
those files are created by this code :
File file = new File("Id.txt");
File file2 = new File("Pass.txt");
File file3 = new File("Remember.txt");
The following three lines will only create file handlers for your program to use. This will not create a file by itself. If you are using the handler to write it will also create a file for you provided you close correctly after writing.
File file = new File("Id.txt");
File file2 = new File("Pass.txt");
File file3 = new File("Remember.txt");
So, a sample code will look like:
File file = new File("Id.txt");
FileWriter fw = new FileWriter(file);
try
{
// write to file
}
finally
{
fw.close();
}
If the file is in the root of your project, this should work:
Path path = Paths.get("foo.txt");
System.out.println(Files.exists(path)); // true
Where exatlcy are the files you want to open in your project?
Please specify the language you use.
Generally you could search the file to see whether the files are in the program bootup folder. For webapps you should pay attention to the "absolute path and the relative path".
=========Edit============
If you are using Jave, then the file should be write out using FileWriter.close() before you can find them in your hard disk.
Ref
Thank you all for your help, I just tried this :
File file = new File("Id.txt");
File file2 = new File("Pass.txt");
File file3 = new File("Remember.txt");
if(file.exists() && file2.exists() && file3.exists()){
// manipulation
}
and it works
I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...
Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.
Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");
The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.
I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.
I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}
Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}
If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.
As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.
Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}
Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.
Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);