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FileReader can not find file even though it is in working directory

I am attempting to read from a file, however the console gives me this error.
Exception in thread "main" java.io.FileNotFoundException: dataEx.txt (The system cannot find the file specified)
This is the code that I am executing.
import java.io.*;
import java.util.*;
public class ReadTest {
public static void main(String[] args) throws IOException {
BufferedReader f = new BufferedReader(new FileReader("dataEx.txt" ));
}
}
This is my project structure
-project
-ReadTest.java
-dataEx.txt

Working directory
Your path is wrong, thus the reader can not find the file. Whereever you think your current working directory should be, that is not where it is.
Execute the following code to know where it is:
System.out.println(Paths.get("").toAbsolutePath());
That is the path to your current working directory. Then compare that result to your expectation. Realize that your expectation was wrong and correct the path to your file or your working directory settings.
It is hard to guess where your directory might be right now. Maybe in your bin folder, next to the .class files. You will see after executing the above code snippet.
NIO
By the way, not sure what exactly you plan on doing with that BufferedReader but you might be interested in the newer modern file API revolving around Files and Paths:
List<String> lines = Files.readAllLines(Paths.get("myFile.txt"));
It also has other neat utility methods for File IO, much better than the cumbersome File class and the clunky BufferedReader.

FileInputStream throws FileNotFoundException

This code throws FileNotFoundException.
Edit: As requested I have included the full StackTrace.
import java.io.FileInputStream;
import java.io.InputStream;
public class ReadFile{
public static void main(String[] args){
InputStream inputstream = new FileInputStream("C:\\file.txt");
}
}
The file "file.txt" is at that location though. I would like to post a screenshot of this as requested, but I can't because I need at least 10 reputation points.

If you are 100% certain that the file exists and you're still getting a FileNotFoundException, than most likely your user or the user running Java has no permission to access this file (since I am using German Windows the dialog is in German, but as you can see "Benutzer" (which is Users) have a denied right to read and execute the file a.txt:
This however, results in a a FileNotFoundException with a localized error message returned :
Exception in thread "main" java.io.FileNotFoundException: C:\a.txt (Zugriff verweigert)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(FileInputStream.java:131)
at java.io.FileInputStream.<init>(FileInputStream.java:87)
at Threadstuff.main(Threadstuff.java:50)
Zugriff verweigert means "access denied". If that isn't the problem either, I guess you should post your full StackTrace.
The other option I mentioned in my comment is an explorer option ("View" -> "Options") in the Folder and Search options -> View:
(roughly translates to "Hide extensions for known extensions")
If this is enabled, the filenames in the explorer are losing their extensions in the view. Meaning that they are shown as "file" instead of "file.txt" - which sometimes leads to the mistake of creating a "file.txt.txt" when renaming a file. And is/was also often used to trick users into thinking they were open a different kind of file (.pdf.exe) - mostly used by bad guys.

Is this really the full Filepath? Better check that one.
Also I'd recommend putting files that are to be read by your program e.g. Textfiles, Images and such into the classpath of your project so when you pack and export it the file paths are not obstructed by being on somebody else's PC where that file does not exist on that path and so on.
This answer suggest you transform the Path of the file to a java conform URL path.

Try the below one.
InputStream inputstream = new FileInputStream("C:"+File.separator+"file.txt");
A better approach would be
File file = new File("C:"+File.separator+"file.txt");
if(file.exists()) {
//Read the file
}
else {
System.out.println("File does not exist);
}
To ensure whether file exists or not in windows, press windows button + r and then paste the file path you have mentioned and after that press enter key. If file is in that location, a notepad with file contents will be opened.

Is there a way to get the file path of the .java file executed or compiled?

In Python the global variable __file__ is the full path of the current file.
System.getProperty("user.dir"); seems to return the path of the current working directory.
I want to get the path of the current .java, .class or package file.
Then use this to get the path to an image.
My project file structure in Netbeans looks like this:
(source: toile-libre.org)
Update to use code suggested from my chosen best answer:
// read image data from picture in package
try {
InputStream instream = TesseractTest.class
.getResourceAsStream("eurotext.tif");
bufferedImage = ImageIO.read(instream);
}
catch (IOException e) {
System.out.println(e.getMessage());
}
This code is used in the usage example from tess4j.
My full code of the usage example is here.

If you want to load an image file stored right next to your class file, use Class::getResourceAsStream(String name).
In your case, that would be:
try (InputStream instream = TesseractTest.class.getResourceAsStream("eurotext.tif")) {
// read stream here
}
This assumes that your build system copies the .tif file to your build folder, which is commonly done by IDEs, but requires extra setup in build tools like Ant and Gradle.
If you package your program to a .jar file, the code will still work, again assuming your build system package the .tif file next to the .class file.

Is there a way to get the file path of the .java file executed or compiled?
For completeness, the literal answer to your question is "not easily and not always".
There is a round-about way to find the source filename for a class on the callstack via StackFrameElement.getFileName(). However, the filename won't always be available1 and it won't necessarily be correct2.
Indeed, it is quite likely that the source tree won't be present on the system where you are executing the code. So if you needed an image file that was stashed in the source tree, you would be out of luck.
1 - It depends on the Java compiler and compilation options that you use. And potentially on other things.
2 - For example, the source tree can be moved or removed after compilation.
Andreas has described the correct way to solve your problem. Make sure that the image file is in your application's JAR file, and access it using getResource or getResourceAsStream. If your application is using an API that requires a filename / pathname in the file system, you may need to extract the resource from the JAR to a temporary file, or something like that.

public class Main {
public static void main(String[] args) throws Exception {
System.out.println(getPackageParent(Main.class, false));
}
public static String getPackageParent(Class<?> cls, boolean include_last_dot)
throws Exception {
StringBuilder sb = new StringBuilder(cls.getPackage().getName());
if (sb.lastIndexOf(".") > 0)
if (include_last_dot)
return sb.delete(sb.lastIndexOf(".") + 1, sb.length())
.toString();
else
return sb.delete(sb.lastIndexOf("."), sb.length()).toString();
return sb.toString();
}
}

java.nio.file Files.move() is copying instead of moving

I'm trying to use Java NIO Files.move method to move a directory. It does copy the directory contents to the new location, but it leaves the old directory in place. I would consider this a copy operation and not a move operation.
Any ideas why this is happening? Here is my code:
Path source = FileSystems.getDefault().getPath("C:\\test-source");
Path destination = FileSystems.getDefault().getPath("C:\\test-destination");
try {
System.out.println("Moving files ...");
Files.move(source, destination, StandardCopyOption.ATOMIC_MOVE);
System.out.println("Done.");
} catch (IOException e) {
System.out.println("Moving failed: " + e.toString());
}
Again, the destination directory appears with all its contents, but the source folder remains in place.

From
this
ATOMIC_MOVE is a file operation.
public static final StandardCopyOption ATOMIC_MOVE
Move the file as an atomic file system operation.
Try StandardCopyOption.REPLACE_EXISTING

It turns out that the code is correct. But the source folder is not being deleted because another process is still working with that folder. When I eliminate the other process (an AWS S3 directory download to the source folder), the move happens as I would expect.

I was encountering this problem with Apache Lucene when trying to commit an IndexWriter. The problem was an IndexSearcher opened on the directory, resulting in a java.nio.file.atomicmovenotsupportedexception. The IndexReader constructor argument was provided by DirectoryReader.open(Directory directory). Changing to use the overload DirectoryReader.open(IndexWriter writer) fixed the problem.

I had this problem on a zipFileSystem but discovered I needed to call fileSystem.close() before it actually got removed from the zip.

Java - FilenotfoundException for reading text file

by running this...
File file = new File("Highscores.scr");
i keep getting this error, and i really don't know how to get around it.
the file is currently sitting in my source packages with my .java files.
I can quite easily read the file by specifying the path but i intend to run this on multiple computers so i need the file to be portable with the program.
this question isnt about reading the text file but rather specifying its location without using an absolute path .
ive searched for the answer but the answers i get are just "specify the name" and "specify the absolute path".
id post an image to make it more clear but i dont have the 10 rep to do so :/
how do i do this?
cheers.

The best way to do this is to put it in your classpath then getResource()
package com.sandbox;
import org.apache.commons.io.FileUtils;
import java.io.File;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Sandbox {
public static void main(String[] args) throws URISyntaxException, IOException {
new Sandbox().run();
}
private void run() throws URISyntaxException, IOException {
URL resource = Sandbox.class.getResource("/my.txt");
File file = new File(resource.toURI());
String s = FileUtils.readFileToString(file);
System.out.println(s);
}
}
I'm doing this because I'm assuming you need a File. But if you have an api which takes an InputStream instead, it's probably better to use getResourceAsStream instead.
Notice the path, /my.txt. That means, "get a file named my.txt that is in the root directory of the classpath". I'm sure you can read more about getResource and getResourceAsStream to learn more about how to do this. But the key thing here is that the classpath for the file will be the same for any computer you give the executable to (as long as you don't move the file around in your classpath).
BTW, if you get a null pointer exception on the line that does new File, that means that you haven't specified the correct classpath for the file.

As far as I remember the default directory with be the same as your project folder level. Put the file one level higher.
-Project/
----src/
----test/
-Highscores.scr

If you are building your code on your eclipse then you need to put your Highscores.scr to your project folder. Try that and check.

You can try to run the following sample program to check which is the current directory your program is picking up.
File f = new File(".");
System.out.println("Current Directory is: " + f.getAbsolutePath());