We Keep Coding

Cannot be resolved to absolute file path because it does not reside in the file system

My Code:
XWPFDocument doc = new XWPFDocument(OPCPackage.open(ResourceUtils.getFile("classpath:assets/OPTIONS_" + jubilar1.getJubiLanguage().toUpperCase() + ".docx")));
I have already tried instead of .getFile(), extractJarFileFromURL or resource.getInputStream() but all this does not work. When I package my project and run it as a jar file and it tries to open the following file it always returns the following message.
Error:
java.io.FileNotFoundException: class path resource [assets/OPTIONS_DE.
docx] cannot be resolved to absolute file path because it does not
reside in the file system:
jar:file:/home/tkf6y/IdeaProjects/hrapps/backend/target/backend-3.0.0.jar!/BOOT-INF/classes!/assets/OPTIONS_EN.docx

So yes it was the problem, as you are now using an InputStream as I suggested. The problem was (and always has been) the getFile stuff. What I suggest to do is don't use what you have now but rather do a new ClassPathResource(your location).getInputStream()) instead, it is easier, or even use a ResourceLoader (a Spring interface you can inject) and then use the path you had an again use getInputStream(). –

This works for me.
String strJson = null;
ClassPathResource classPathResource = new ClassPathResource("json/data.json");
try {
byte[] binaryData = FileCopyUtils.copyToByteArray(classPathResource.getInputStream());
strJson = new String(binaryData, StandardCharsets.UTF_8);
} catch (IOException e) {
e.printStackTrace();
}

URI not hierarchical need to use File class for a method

I need to open a video file with my code, and it works perfectly fine in Eclipse but when I export into a runnable JAR, i get an error "URI not hierarchical".
I have seen people suggest using getResourceAsStream(), but i need to have a file object as i am using Desktop.getDesktop.open(File). Can anyone help me out?
Here is the code:
try {
URI path1 = getClass().getResource("/videos/tutorialVid1.mp4").toURI();
File f = new File(path1);
Desktop.getDesktop().open(f);
} catch (Exception e) {
e.printStackTrace();
}
if it helps my folder list is like
Src
videos
videoFile.mp4
EDIT:
I plan to run this on windows only, and use launch4j to create an exe.

You can copy the file from the jar to a temporary file and open that.
Here's a method to create a temporary file for a given jar resource:
public static File createTempFile(String path) {
String[] parts = path.split("/");
File f = File.createTempFile(parts[parts.length - 1], ".tmp");
f.deleteOnExit();
try (Inputstream in = getClass().getResourceAsStream(path)) {
Files.copy(in, f.toPath(), StandardCopyOption.REPLACE_EXISTING);
}
return f;
}
And here's an example of how you'd use it:
Desktop.getDesktop().open(createTempFile("/videos/tutorialVid1.mp4"));

renaming file name inside a zip file

trying to rename internal file within a zip file without having to extract and then re-zip programatically.
example. test.zip contains test.txt, i want to change it so that test.zip will contain newtest.txt(test.txt renamed to newtest.txt, contents remain the same)
came across this link that works but unfortunately it expects test.txt to exist on the system. In the example the srcfile should exist on the server.
Blockquote Rename file in zip with zip4j
Then icame across zipnote on Linux that does the trick but unfortunately the version i have doesnt work for files >4GB.
Any suggestions on how to accomplish this? prefereably in java.

This should be possible using Java 7 Zip FileSystem provider, something like:
// syntax defined in java.net.JarURLConnection
URI uri = URI.create("jar:file:/directoryPath/file.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.<String, Object>emptyMap())) {
Path sourceURI = zipfs.getPath("/pathToDirectoryInsideZip/file.txt");
Path destinationURI = zipfs.getPath("/pathToDirectoryInsideZip/renamed.txt");
Files.move(sourceURI, destinationURI);
}

Using zip4j, I am modifying and re-writing the file headers inside of the central directory section to avoid rewriting the entire zip file:
ArrayList<FileHeader> FHs = (ArrayList<FileHeader>) zipFile.getFileHeaders();
FHs.get(0).setFileName("namename.mp4");
FHs.get(0).setFileNameLength("namename.mp4".getBytes("UTF-8").length);
zipFile.updateHeaders ();
//where updateHeaders is :
public void updateHeaders() throws ZipException, IOException {
checkZipModel();
if (this.zipModel == null) {
throw new ZipException("internal error: zip model is null");
}
if (Zip4jUtil.checkFileExists(file)) {
if (zipModel.isSplitArchive()) {
throw new ZipException("Zip file already exists. Zip file format does not allow updating split/spanned files");
}
}
long offset = zipModel.getEndCentralDirRecord().getOffsetOfStartOfCentralDir();
HeaderWriter headerWriter = new HeaderWriter();
SplitOutputStream splitOutputStream = new SplitOutputStream(new File(zipModel.getZipFile()), -1);
splitOutputStream.seek(offset);
headerWriter.finalizeZipFile(zipModel, splitOutputStream);
splitOutputStream.close();
}
The name field in the local file header section remains unchanged, so there will be a mismatch exception in this library.
It's tricky but maybe problematic, I don't know..

set directory to a path in a file

I just want to set the directory to a path I have written in a file before.
Therefore I used :
fileChooser.setCurrentDirectory(new File("path.txt"));
and in path.txt the path is given. But unfortunately this does not work out and I wonder why :P.
I think I got it all wrong with the setCurrentDic..

setCurrentDirectory takes a file representing a directory as parameter. Not a text file where a path is written.
To do what you want, you have to read the file "path.txt", create a File object with the contents that you just read, and pass this file to setCurrentDirectory :
String pathWrittenInTextFile = readFileAsString(new File("path.txt"));
File theDirectory = new File(pathWrittenInTextFile);
fileChooser.setCurrentDirectory(theDirectory);

You have to read the contents of path.txt. Thea easiest way is through commons-io:
String fileContents = IOUtils.toString(new FileInputStream("path.txt"));
File dir = new File(fileContents);
You can also use FileUtils.readFileToString(..)

JFileChooser chooser = new JFileChooser();
try {
// Create a File object containing the canonical path of the
// desired directory
File f = new File(new File(".").getCanonicalPath());
// Set the current directory
chooser.setCurrentDirectory(f);
} catch (IOException e) {
}

Java, reading a file from current directory?

I want a java program that reads a user specified filename from the current directory (the same directory where the .class file is run).
In other words, if the user specifies the file name to be "myFile.txt", and that file is already in the current directory:
reader = new BufferedReader(new FileReader("myFile.txt"));
does not work. Why?
I'm running it in windows.

Try
System.getProperty("user.dir")
It returns the current working directory.

The current directory is not (necessarily) the directory the .class file is in. It's working directory of the process. (ie: the directory you were in when you started the JVM)
You can load files from the same directory* as the .class file with getResourceAsStream(). That'll give you an InputStream which you can convert to a Reader with InputStreamReader.
*Note that this "directory" may actually be a jar file, depending on where the class was loaded from.

None of the above answer works for me. Here is what works for me.
Let's say your class name is Foo.java, to access to the myFile.txt in the same folder as Foo.java, use this code:
URL path = Foo.class.getResource("myFile.txt");
File f = new File(path.getFile());
reader = new BufferedReader(new FileReader(f));

Files in your project are available to you relative to your src folder. if you know which package or folder myfile.txt will be in, say it is in
----src
--------package1
------------myfile.txt
------------Prog.java
you can specify its path as "src/package1/myfile.txt" from Prog.java

If you know your file will live where your classes are, that directory will be on your classpath. In that case, you can be sure that this solution will solve your problem:
URL path = ClassLoader.getSystemResource("myFile.txt");
if(path==null) {
//The file was not found, insert error handling here
}
File f = new File(path.toURI());
reader = new BufferedReader(new FileReader(f));

Thanks #Laurence Gonsalves your answer helped me a lot.
your current directory will working directory of proccess so you have to give full path start from your src directory like mentioned below:
public class Run {
public static void main(String[] args) {
File inputFile = new File("./src/main/java/input.txt");
try {
Scanner reader = new Scanner(inputFile);
while (reader.hasNextLine()) {
String data = reader.nextLine();
System.out.println(data);
}
reader.close();
} catch (FileNotFoundException e) {
System.out.println("scanner error");
e.printStackTrace();
}
}
}
While my input.txt file is in same directory.

Try this:
BufferedReader br = new BufferedReader(new FileReader("java_module_name/src/file_name.txt"));

try using "."
E.g.
File currentDirectory = new File(".");
This worked for me