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prunsrv service with external properties file

I have a java .jar file I created using NetBeans. I am using apaches procrun (prunsrv.exe) to install that .jar as a Windows Service. I modified the code to get a property from a config.properties file.
I added the config.properties file to the same folder that my .jar file resides in.
My code is as follows:
Properties props = new Properties();
InputStream inputStream = MyService.class.getClassLoader().getResourceAsStream("config.properties");
props.load(inputStream);
On the last line of my code, I am getting a NPE when I attempt to start my service. I assume this is because the file is not found.
I modified the manifest.mf as follows:
Class-Path: .
I also tried copying config.properties to the "lib" folder (subfolder to where my .jar file is located). Same results.
I modified the "set PR_CLASSPATH" line in the batch file that installs the service as follows:
set PR_CLASSPATH=MyService.jar;.
Still same NPE.
How can I get my code to recognize my config.properties file once the service has been installed?
Thanks,
Raymond

This is what I use to load resources in these situations and seems to work most of the time:
public static InputStream getResourceAsStream(String path) {
return Thread.currentThread().getContextClassLoader().getResourceAsStream(path);
}
Could you check, if this helps in your case?
Another strategy to overcome this problem is the one described in my comment:
"copy the configuration file to a specific absolute folder path (i.e. c:\test) and change the classpath to point to that folder (set PR_CLASSPATH=MyService.jar;.;c:\test)"

FileNotFoundException while running as a jar

FileInputStream fstream = new FileInputStream("abc.txt")
is throwing a FileNotFoundExceptionn while running as a jar. Why ? Normally it is able to find while running from main method.

class MyClass{
InputStream fstream = this.getClass().getResourceAsStream("abc.txt");
}
This code should be used.
And the files(in this case abc.txt) should be kept , in the Object references class location. That means , this.getClass refers to the location of some folder i.e, com/myfolder/MyClass.java folder .
So we should keep the abc.txt in com/myfolder this location.

If your file is packaged with your jar then you should to get information using getClass().getResource(url):
FileInputStream inputStream =
new FileInputStream(new File(getClass().getResource(/path/to/your/file/abc.txt).toURI()));
Else you need to create it always in the same path with your jar and you can get it like you do :
src/myJar.jar
src/folder/abc.txt
FileInputStream fstream = new FileInputStream("folder/abc.txt");
You can read here also :
How do I load a file from resource folder? and File loading by getClass().getResource()

You can use FileInputStream only when you actually have a file on the computer's filesystem. When you package your text file in the jar file for your program, it is not a file in the filesystem. It is an entry inside the jar file.
The good news is that it is even easier, in Java, to access the file this way: it is in your classpath, so you can use getResourceAsStream().
InputStream stream = getClass().getResourceAsStream("abc.txt");
If you have your classpath set up correctly, this will work regardless of whether it is a file in a directory (such as during development), or an entry in a jar file (such as when released).

It's because your working directory will probably be different under the two environments. Try adding the line
System.out.println(new File("abc.txt").getAbsolutePath());
to see where it is actually looking for the file.

ClassLoader shows wrong path on Linux

I am using classLoader to load xml files located under /src/XMLS :
String m_path = "XMLS/file.xml"
ClassLoader cl = getClass.getClassLoader();
file f1 = new file(cl.getResource(m_path).getFile));
Running on windows it works fine but after export to jar and running it on Linux I get FileNotFoundException - /XMLS/file.xml.
I had tried this solutions and I dont think that the problem is in the read from the .jar file. any other ideas for what I am doing wrong?

This will not work for a resource inside a jar file, which is not a file on the filesystem. Instead, you need to use getResourceAsStream(), which returns an InputStream to use directly:
InputStream in = someClass.getClassLoader().getResourceAsStream("/XMLS/file.xml");

How can i find the path of my file in my Java project file?

I put a file inside my Java project file and i want to read it but how can i find the path name with Java.
Here i put it in C driver but i just want to find path by just writing the name of file. Is there a function for it?
FileInputStream fstream1 = new FileInputStream("C:/en-GB.dic");

If the file is inside the jar file generated for your project (or in the classpath used by your project, generally), under the package com.foo.bar, you can load it using
SomeClassOfYourProject.class.getResourceAsStream("/com/foo/bar/en-GB.dic");
If it's not in the classpath, and you launch the application (using java.exe) from the directory c:\baz, and the file is under c:\baz\boom\, the you can load it using
new FileInputStream("boom/en-GB.dic");

Place it in your classpath, If it is a web application WEB-INF/classes/yourfile.ext, if it is a standalone application place it in bin directory of your application (default class directory is bin).
Then you could read by using one of the following ways.
InputStream in = this.getClass().getClassLoader().getResourceAsStream("yourfile.ext");
Or
InputStream in = this.getClass().getResourceAsStream("/yourfile.ext");
You can read online for the differences between above two approaches.

Executable .jar fails to start because of .properties file loading

I have the problem running executable .jar file. I've created a project which contains a .properties file. It works just fine when I start it from eclipse, but when I export it to executable .jar file and try to run it with:
java -jar myfile.jar
I get the following exception:
(couldn't post image here)
http://imageshack.us/photo/my-images/824/29583616.png/
I've checked my manifest file in the .jar and it contains the
Class-Path: .
And here's the properties file loading:
properties = new Properties();
properties.load(new FileInputStream(
"src/com/resources/treeView.properties"));
Any idea what causes this exception?

If the properties file is inside the jar file, you cannot access it as a file.
You need to ask the classloader to get the resource as an inputstream. See Getting the inputstream from a classpath resource (XML file)

In Eclipse (and in most IDEs) the current directory is the project's root directory. This means that Class-Path: . means something else in Eclipse than when you run it from the command line. This is why you wrote "src/com/...". Remove "src":
properties.load(new FileInputStream("com/resources/treeView.properties"));

Your properties file is within JAR file. So, use : ClassLoader.getResourceAsStream().