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Error message: The file "countries.geo.json" is missing or inaccessible

I am taking the Coursera OOP in Java class. In the module 4 assignment, I run the code that the course provides in EarthquakeCityMap.java,
and I get an error as "The file "countries.geo.json" is missing or inaccessible, make sure the URL is valid or that the file has been added to your sketch and is readable.
Exception in thread "Animation Thread" java.lang.NullPointerException"
I tried to set countryFile as
"../data/countries.geo.json",
"data/countries.geo.json",
and the complete path of countries file,
but still didn't solve the problem.
//this error points to the code
private String countryFile = "countries.geo.json";
List<Feature> countries = GeoJSONReader.loadData(this, countryFile);"
//the countries file is saved in data folder.
Poject folder listing

"countries.geo.json" (unless changed in GeoJSONReader manipulates this path) will be relative to the compiled java .class files in the IntelliJ's project out folder.
If this in GeoJSONReader.loadData(this, countryFile); is a PApplet instance you can use sketchPath() to make that path relative to the folder from which the sketch runs:
List<Feature> countries = GeoJSONReader.loadData(this, this.sketchPath("data"+File.separator+countryFile));
The above snippet is based on an assumption so the syntax in your code might be slightly different, but hopefully this illustrates how you'd use sketchPath().
Additionally there's a dataPath() as well which you can test from your main PApplet in setup() as a test:
String fullJSONPath = dataPath("countries.geo.json");
println("fullJSONPath: " + fullJSONPath);//hopefully this prints the full path to the json file on your machine
println(new File(fullJSONPath).exists());//hopefully this prints true

If you specified the full path and it didn’t work, you probably forgot to escape the \ character with another backslash. The backslash character is special and needs to be doubled for windows path to be interpreted properly. For instance “c:\\users\\...”. You can also specify / instead of \ and it would work : “c:/users/...”
That said, the path resolution of a file when relative (IE not being absolute to the file system root) is relative to the working directory of the executed app. Typically, in an IDE without any special configuration, the working directory would be the root path of the project. So in order to get the relative file path resolved properly, you would have to specify the path as “data/countries.geo.json”.
You can also find out what path you are in when you run the app by doing a System.out.println(new java.io.File(“.”).getAbsolutePath()) and craft the relative path according to this folder.

right way to make a jar file from javafx application

I have a javafx project, which contains multiple paths for images and text files :
private Image imgMan = new Image(getClass().getResource("../man.gif").toExternalForm());
FileHelper.resetScores("./bin/application/MAP/BestScores.txt");
...
When i launch from eclipse, it work normally, and access to images and files without any problem.
But when i try to export my project to a jar file, it export correctly, but it don't launch !
I try to launch it from cmd, the trace of stack said that he don't know the paths...
Caused by: java.lang.NullPointerException
at application.Client.(Client.java:31)
(line 31 in my code refer to the first line of code given in the question)
I try to create a resource folder and put all files into it, but no result.
So what is the best way to make it ?
where must i create the resource folder ?
and how to access the files into it from the code ?
Thank you

There are several thing you should check:
verify the path in your jar against the class you are looking. Your image must be there.
verify you have successfully loaded a resource because using it, eg: check if getResource returns null.
For the first point, it depends on how you build your jar:
Eclipse will by default copy class file and resources to bin unless you use m2e. If you use the Extract runnable JAR (from File > Export menu), it may ignore some resources.
If you use Maven then your images must be in src/main/resources by default.
For the second point, you should use a method that should check the resource exists before delegating to Image. While it won't change your core problem, you would have a less subtile error:
static javafx.scene.image.Image loadImage(Class<?> source, String path) {
final InputStream is = source.getResourceAsStream(path);
if (null == is) {
throw new IllegalStateException("Could not load image from " + source + " path: " + path);
}
try (is) { // Java 9 -> you may want to use InputStream is2 = is
return new javafx.scene.image.Image(is); // use is2 for Java < 9
}
}
You should also try with an absolute path (from the root of the jar, or your src/main/resources if you use maven):
Image image = loadImage(this.getClass(), "/images/man.gif");

File path issue on windows vs Unix in java

In my program, I am reading a resource file for a unit test. I use file path as:
\\\path\\\to\\\file
On my machine(Windows) this runs fine. But on server(Unix), this fails, and I have to change it to: /path/to/file
But Java is supposed to be platform independent. So isn't this behaviour unexpected?

Use FileSystem.getSeparator() or System.getProperty("file.separator") instead of using slashes.
EDIT:
You can get an instance of FileSystem via FileSystems.getDefault (JDK 1.7+)

You can use File.separator to get the appropriate character in a platform-independent way.

Java is platform independent. The file path-es and some system calls are not.
As long as the path is relative, you can use File.separator:
String path = "path" + File.separator + "to" + File.separator + "file";
System.out.println(path); // prints path\to\file on windows
Sometimes it's an option is to provide a Properties file and let the user define path of that actual file. This way full paths are okay too. You can read the properties like this:
Properties props = new Properties();
props.load(new FileInputStream(filePath));
The next question is: how to specify the location of that file? That might be either a file on a relative path. If that's not viable for your app, then you can let the user specify it in a system property:
java ... -DconfigFile=C:\TEMP\asd.txt .... -jar myapp.jar
Then you can access it like this:
// prints C:\TEMP\asd.txt if you specified -DconfigFile=C:\TEMP\asd.txt
System.out.println(System.getProperty("configFile"));

This is the expected behaviour.
Java code compiles on any machine/OS provided you have the right version of Java installed on it.
However, at run time, your code sees only a variable value like another one, which happens to be \path\to\file
When it talks to the file system, it uses that particular value ; the file system then tries to find that path you've given to it ; which is why one syntax works fine on Windows but will not work on Linux.

Better way of doing this is :
val pathUri = Paths.get(".//src//test//res//file.txt").toUri()
val is = FileInputStream((File(pathUri)))

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();
}
}

Get running file name in Java?

Is it possible to get the file name which is being executed. Like e.g. If i am running a Jar file, and inside i want to add code functionality that can detect the file name so that if this jar is renamed, code should be able to detect that.
Is there any possibility of doing this without scanning the files in current directory?

Try this:
String path = Test.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath = URLDecoder.decode(path, "UTF-8");
http://www.rgagnon.com/javadetails/java-0300.html
http://docs.oracle.com/javase/7/docs/api/java/lang/Class.html#getProtectionDomain()
http://docs.oracle.com/javase/7/docs/api/java/security/ProtectionDomain.html

you could check if System.getEnv() has some value indicating which file was invoked