Get full command line from Java [duplicate]

Question

How can I from within Java code find out which and how the running JVM was launched? My code shall spawn another Java process hence I'd be especially interested in the first parameter - which usually (in C, Pascal, Bash, Python, ...) points to the currently running executable.

So when a Java application is run like

d:\openjdk\bin\java -Xmx500g -Dprop=name -jar my.jar param1 param2

I can access command line parameters in my main method like

public class Main {
    public static void main(String[] args) {
        System.out.println("main called with " + args);
    }
}

but that will deliver access to param1 and param2 only. How would I get the full command line?

Answer 1

Following Determining location of JVM executable during runtime I found the real answer to be:

ProcessHandle.current().info().commandLine().orElseThrow();

It returns the full command line as perceived by the operating system and thus contains all information: executable, options, main class, arguments, ...

Thank you, @XtremeBaumer

Answer 2

To get the command, use ProcessHandle adapted from your answer:

String command = ProcessHandle
                 .current()
                 .info()
                 .command()
                 .orElse("<unable to determine command>"); // .orElseThrow()  to get Exception

---

Use the Runtime Management Bean RuntimeMXBean to obtain the VM arguments and the class path, like in:

RuntimeMXBean mxBean = ManagementFactory.getRuntimeMXBean();
String classPath = mxBean.getClassPath();
List<String> inputArgs = mxBean.getInputArguments();

This will return all arguments passed to the virtual machine, not the parameters passed to the main method.

---

The code source can be get from the ProtectionDomain of the class.