Question

I’m making an editor-like program. If the user chooses File->Open in the main window I want to start a new copy of the editor process with the chosen filename as an argument. However, for that I need to know what command was used to start the first process:

java -jar myapp.jar blabalsomearguments // --- need this information
> Open File (fileUrl)
> exec("java -jar myapp.jar blabalsomearguments fileUrl");

I’m not looking for an in-process solution, I’ve already implemented that. I’d like to have the benefits that seperate processes bring.

Answer 1

Since you are launching Java -> Java, you can use the existing classpath to set the classpath on the command line. This type of thing works really nice in the dev environment too.

ProcessBuilder selfLauncher = new ProcessBuilder(
 "java", "-cp", System.getProperty("java.class.path"),
 "com.my.mainClass" );
selfLauncher.start();

Update:
For executable jar files, you will have a classpath which is simply the relative path to the jar file itself. If you want the command line arguments, you will have to save them from main, and re-apply them when launching.

You can see this by packing the following program into a jar. I’m not actually sure what happens if you have jars inside the executable jar file. They probably show up in the classpath.

public class TestJarPath {
    public static void main(String args[]) throws Exception {
        for (String s : args)
         System.out.print("[" + s + "] ");
        System.out.println();

        String cp = System.getProperty("java.class.path");
        for (String s : cp.split(";"))
          System.out.println(s);
    } 
}

For java -jar ..\tst.jar X, you get output like:

[X]
..\tst.jar