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Sound will not play in JAR file

I have a problem and I hope you can help me.
Some talk about what I am doing so you know what's going on: So at the moment I'm trying to program a litte piece of software which can play me some music files (mp3 files to be exact, so i'm using the jLayer API). I'm working with Netbeans and I have succesfully imported a music file in the project. If I build my program and open the resulting JAR file with an archive program, I can find my music file in there. My function which I'm using goes like this:
public static String play(String file) {
File test = new File(file);
try {
FileInputStream in = new FileInputStream(test);
Player pl = new Player(in);
pl.play();
return "success";
}
catch (Exception e) {
return e.toString();
}
}
As you can see I'm getting a String with the Path Name and refactor him so I can play the file. I'm calling the function with the following code (the music file is saved in the ressources package):
MP3.play(getClass().getResource("/ressources/angel.mp3").getPath())
So if I start the programm via Netbeans everything works perfectly fine. But if I create a JAR file and start the program nothing happens. The Exception getting is the following:
java.io.FileNotFoundException: file:\C:\Users\Raphael\Documents\NetBeansProjects\MP3\dist\MP3.jar!\ressources\angel.mp3
It says the File does not exist but if I check my JAR the file is there......
Another strange thing I found out is the following: If I use the following function to play the music file:
public static String play(InputStream test) {
try {
Player pl = new Player(test);
pl.play();
return "success";
}
catch (Exception e) {
return e.toString();
}
}
and call the function with the following argument:
MP3.play(getClass().getResourceAsStream("/ressources/angel.mp3"));
Everything works fine in both Netbeans and the final JAR. Can anybody explain me what I'm doing wrong and only the second function works in the JAR version?
It would be really nice if you could help me in this matter.
Greetings,
xXKnightRiderXx

I am assuming that you have 2 packages 1 is src where your .java files is located and other is resources where your sound files is located
So i suggest you to use
MP3.play(getClass().getResourceAsStream("/angel.mp3"));
Because GetResource() automatically finds the resource package

In Java, why does class.getResource("/path/to/res"); not work in the runnable jar when copying files to the system?

I have been working on a project that requires the user to "install" the program upon running it the first time. This installation needs to copy all the resources from my "res" folder to a dedicated directory on the user's hard drive. I have the following chunk of code that was working perfectly fine, but when I export the runnable jar from eclipse, I received a stack trace which indicated that the InputStream was null. The install loop passes the path of each file in the array list to the export function, which is where the issue is (with the InputStream). The paths are being passed correctly in both Eclipse and the runnable jar, so I doubt that is the issue. I have done my research and found other questions like this, but none of the suggested fixes (using a classloader, etc) have worked. I don't understand why the method I have now works in Eclipse but not in the jar?
(There also exists an ArrayList of File called installFiles)
private static String installFilesLocationOnDisk=System.getProperty("user.home")+"/Documents/[project name]/Resources/";
public static boolean tryInstall(){
for(File file:installFiles){
//for each file, make the required directories for its extraction location
new File(file.getParent()).mkdirs();
try {
//export the file from the jar to the system
exportResource("/"+file.getPath().substring(installFilesLocationOnDisk.length()));
} catch (Exception e) {
return false;
}
}
return true;
}
private static void exportResource(String resourceName) throws Exception {
InputStream resourcesInputStream = null;
OutputStream resourcesOutputStream = null;
//the output location for exported files
String outputLocation = new File(installFilesLocationOnDisk).getPath().replace('\\', '/');
try {
//This is where the issue arises when the jar is exported and ran.
resourcesInputStream = InstallFiles.class.getResourceAsStream(resourceName);
if(resourcesInputStream == null){
throw new Exception("Cannot get resource \"" + resourceName + "\" from Jar file.");
}
//Write the data from jar's resource to system file
int readBytes;
byte[] buffer = new byte[4096];
resourcesOutputStream = new FileOutputStream(outputLocation + resourceName);
while ((readBytes = resourcesInputStream.read(buffer)) > 0) {
resourcesOutputStream.write(buffer, 0, readBytes);
}
} catch (Exception ex) {
ex.printStackTrace();
System.exit(1);
} finally {
//Close streams
resourcesInputStream.close();
resourcesOutputStream.close();
}
}
Stack Trace:
java.lang.Exception: Cannot get resource "/textures\gameIcon.png" from Jar file.
All help is appreciated! Thanks

Stack Trace:
java.lang.Exception: Cannot get resource "/textures\gameIcon.png" from Jar file.
The name if the resource is wrong. As the Javadoc of ClassLoader.getResource(String) describes (and Class.getResourceAsStream(String) refers to ClassLoader for details):
The name of a resource is a /-separated path name that identifies
the resource.
No matter whether you get your resources from the File system or from a Jar File, you should always use / as the separator.
Using \ may sometimes work, and sometimes not: there's no guarantee. But it's always an error.
In your case, the solution is a change in the way that you invoke exportResource:
String path = file.getPath().substring(installFilesLocationOnDisk.length());
exportResource("/" + path.replace(File.pathSeparatorChar, '/'));

Rename your JAR file to ZIP, uncompress it and check where did resources go.
There is a possibility you're using Maven with Eclipse, and this means exporting Runnable JAR using Eclipse's functionality won't place resources in JAR properly (they'll end up under folder resources inside the JAR if you're using default Maven folder names conventions).
If that is the case, you should use Maven's Assembly Plugin (or a Shade plugin for "uber-JAR") to create your runnable JAR.
Even if you're not using Maven, you should check if the resources are placed correctly in the resulting JAR.
P.S. Also don't do this:
.getPath().replace('\\', '/');
And never rely on particular separator character - use java.io.File.separator to determine system's file separator character.

File not found exception with external files

Hi i have made a small program that reads a config file. This file is stored outside the actual jar file. On the same level as the jarfile actually.
When i start my program from a commandline in the actual directory (ie. D:\test\java -jar name.jar argument0 argument1) in runs perfectly.
But when i try to run the program from another location then the actual directory i get the filenotfound exception (ie. D:\java -jar D:\test\name.jar argument0 argument1).
The basic functionality does seem to work, what am i doing wrong?
As requested a part of the code:
public LoadConfig() {
Properties properties = new Properties();
try {
// load the properties file
properties.load(new FileInputStream("ibantools.config.properties"));
} catch (IOException ex) {
ex.printStackTrace();
} // end catch
// get the actual values, if the file can't be read it will use the default values.
this.environment = properties.getProperty("application.environment","tst");
this.cbc = properties.getProperty("check.bankcode","true");
this.bankcodefile = properties.getProperty("check.bankcodefile","bankcodes.txt");
} // end loadconfig
The folder looks like this:
This works:
This doesn't:
The jar doesn't contain the text file.

When reading a File using the String/path constructors of File, FileInpustream, etc.. a relative path is derived from the working directory - the directory where you started your program.
When reading a file from a Jar, the file being external to the jar, you have at least two options :
Provide an absolute path: D:/blah/foo/bar
Make the directory where your file is located part of the class path and use this.getClass().getClassLoader().getResourceAsStream("myfile")
The latter is probably more appropriate for reading configuration files stored in a path relative to the location of your application.

There could be one more possibility:
If one part of your code is writing the file and another one is reading, then it is good to consider that the reader is reading before the writer finishes writing the file.
You can cross check this case by putting your code on debug mode. If it works fine there and gives you FileNotFoundException, then surely this could be the potential reason of this exception.
Now, how to resolve:
You can use retry mechanism something similar to below code block
if(!file..exists()){
Thread.sleep(200);
}
in your code and change the sleep value according to your needs.
Hope that helps.!!

Java: get absolute path of project

I'm trying to run a exe file in path outside of the current package. My code.java file that runs it is in
%Workspace_path%\Project\src\main\java\com\util\code.java
However the directory of where the exe is
%Workspace_path%\Project\src\main\resources\program.exe
If possible, it seems like the best solution here would be to get the absolute path of the Project then append "src\main\resources\" to it. Is there a good way to do this or is there an alternative solution?
I'm using Eclipse, but it would great if it could be used in other IDEs too. Thanks for any help.

The de facto approach to solving this is to bundle the EXE as a classpath resource. It seems you have arranged for this already.
When working with classpath resources, a mature program should not assume that the resource is in the filesystem. The resources could be packaged in a JAR file, or even in a WAR file. The only thing you can trust at that point is the standard methods for accessing resources in Java, as hinted below.
The way to solve your problem, then, is to access the resource contents using the de facto standard of invoking Class.getResourceAsStream (or ClassLoader.getResourceAsStream), save the contents to a temporary file, and execute from that file. This will guarantee your program works correctly regardless of its packaging.
In other words:
Invoke getClass().getResourceAsStream("/program.exe"). From static methods, you can't call getClass, so use the name of your current class instead, as in MyClass.class.getResourceAsStream. This returns an InputStream.
Create a temporary file, preferably using File.createTempFile. This returns a File object identifying the newly created file.
Open an OutputStream to this temp file.
Use the two streams to copy the data from the resource into the temp file. You can use IOUtils.copy if you're into Apache Commons tools. Don't forget to close the two streams when done with this step.
Execute the program thus stored in the temporary file.
Clean up.
In other words (code snippet added later):
private void executeProgramFromClasspath() throws IOException {
// Open resource stream.
InputStream input = getClass().getResourceAsStream("/program.exe");
if (input == null) {
throw new IllegalStateException("Missing classpath resource.");
}
// Transfer.
OutputStream output = null;
try {
// Create temporary file. May throw IOException.
File temporaryFile = File.createTempFile(getClass().getName(), "");
output = new FileOutputStream(temporaryFile);
output = new BufferedOutputStream(output);
IOUtils.copy(input, output);
} finally {
// Close streams.
IOUtils.closeQuietly(input);
IOUtils.closeQuietly(output);
}
// Execute.
try {
String path = temporaryFile.getAbsolutePath();
ProcessBuilder processBuilder = new ProcessBuilder(path);
Process process = processBuilder.start();
process.waitFor();
} catch (InterruptedException e) {
// Optional catch. Keeps the method signature uncluttered.
throw new IOException(e);
} finally {
// Clean up
if (!temporaryFile.delete()) {
// Log this issue, or throw an error.
}
}
}

Well,in your context,the project root is happen to be the current path
.
,that is where the java.exe start to execute,so a easy way is:
String exePath="src\\main\\resources\\program.exe";
File exeFile=new File(".",exePath);
System.out.println(exeFile.getAbusolutePath());
...
I tested this code on Eclipse,It's ok. I think is should work on different ide.
Good Luck!

Locating default root folder for an Eclipse project

I am reading a file as follows:
File imgLoc = new File("Player.gif");
BufferedImage image = null;
try {
image = ImageIO.read(imgLoc);
}
catch(Exception ex)
{
System.out.println("Image read error");
System.exit(1);
}
return image;
I do not know where to place my file to make the Eclipse IDE, and my project can detect it when I run my code.
Is there a better way of creating a BufferedImage from an image file stored in your project directory?

Take a look in the comments for Class.getResource and Class.getResourceAsStream. These are probably what you really want to use as they will work whether you are running from within the directory of an Eclipse project, or from a JAR file after you package everything up.
You use them along the lines of:
InputStream in = MyClass.class.getResourceAsStream("Player.gif");
In this case, Java would look for the file "Player.gif" next to the MyClass.class file. That is, if the full package/class name is "com.package.MyClass", then Java will look for a file in "[project]/bin/com/package/Player.gif". The comments for getResourceAsStream indicate that if you lead with a slash, i.e. "/Player.gif", then it'll look in the root (i.e. the "bin" directory).
Note that you can drop the file in the "src" directory and Eclipse will automatically copy it to the "bin" directory at build time.

In the run dialog you can choose the directory. The default is the project root.

From my experience it seems to be the containing projects directory by default, but there is a simple way to find out:
System.out.println(new File(".").getAbsolutePath());

Are you trying to write a plugin for Eclipse or is it a regular project?
In the latter case, wouldn't that depend on where the program is installed and executed in the end?
While trying it out and running it from Eclipse, I'd guess that it would find the file in the project workspace. You should be able to find that out by opening the properties dialog for the project, and looking under the Resource entry.
Also, you can add resources to a project by using the Import menu option.

The default root folder for any Eclipse project is also a relative path of that application.
Below are steps I used for my Eclipse 4.8.0 and Java 1.8 project.
I - Place your file you want to interact with along the BIN and SRS folders of your project and not in one of those folders.
II - Implement below code in your main() method.
public static void main(String [] args) throws IOException {
FileReader myFileReader;
BufferedReader myReaderHelper;
try {
String localDir = System.getProperty("user.dir");
myFileReader = new FileReader(localDir + "\\yourFile.fileExtension");
myReaderHelper = new BufferedReader(myFileReader);
if (myReaderHelper.readLine() != null) {
StringTokenizer myTokens =
new StringTokenizer((String)myReaderHelper.readLine(), "," );
System.out.println(myTokens.nextToken().toString()); // - reading first item
}
} catch (FileNotFoundException myFileException) {
myFileException.printStackTrace(); } } // End of main()
III - Implement a loop to iterate through elements of your file if your logic requires this.