Find absolute java.exe path programmatically from java code

Question

If I have a java jar or class file which is launched by the user (assuming java path is set in environment variables), so how can i from within the code, figure out absolute path of java.exe/javaw.exe from which this file is being launched.

Like on ubuntu we can run: % which java and it shows the path.

However on windows, if i check System.getenv() it may happen that there are multiple path's found e.g for old or new version. If through cmd line, I run java -version it does not show the path.

Can you tell me either through pure java or command line on windows how is it possible to find out the location of javaw.exe?

Answer 1

String javaHome = System.getProperty("java.home");

---

Can you tell me either through pure Java ... on windows how is it possible to find out the location of javaw.exe?

E.G.

import java.io.File;

class JavawLocation {

    public static void main(String[] args) {
        String javaHome = System.getProperty("java.home");
        File f = new File(javaHome);
        f = new File(f, "bin");
        f = new File(f, "javaw.exe");
        System.out.println(f + "    exists: " + f.exists());
    }
}

Output

C:\Program Files (x86)\Java\jdk1.6.0_29\jre\bin\javaw.exe    exists: true
Press any key to continue . . .

And yes, I am confident that will work in a JRE.

Answer 2

an issue with using "System.getProperty("java.home");", is that it is not always the java exe that the jar is running on, if you want to get that, you can use "System.getProperty("sun.boot.library.path");", from there you can find "java", "java.exe", "javaw", or "javaw.exe"... However there is still an issue with this, java will run just fine if the executable has been renamed, and the actual java executable's structure changes from different JRE's/JDKS's, so there is not much way to find the java exe if it has been renamed. unless someone else has a method ofc, in which case, can you share? :)

(Also, I have seen some people suggest using the first index of System.getProperty("java.library.path");, note, this might not work if the user/launcher has manually set the library path, something which is not too uncommon)

Answer 3

If you are running on Java 9+ you can use following snippet to check the Java command:

import java.io.IOException;
import java.util.Optional;

public class ProcessUtils {
  private ProcessUtils() {}

  public static Optional getCommandOfCurrentProcess() {
     ProcessHandle processHandle = ProcessHandle.current();
     return processHandle.info().command();
  }

  public static void main(String[] args) throws IOException {
    System.out.println(ProcessUtils.getCommandOfCurrentProcess());
  }
}

This will use the new Java Process API to get the handle of itself. From there you can parse the path as needed.

Source

Answer 4

Here's a slightly more generalised solution that I came up with. Maybe useful:

private static String javaExe()
{
    final String JAVA_HOME = System.getProperty("java.home");
    final File BIN = new File(JAVA_HOME, "bin");
    File exe = new File(BIN, "java");

    if (!exe.exists())
    {
        // We might be on Windows, which needs an exe extension
        exe = new File(BIN, "java.exe");
    }

    if (exe.exists())
    {
        return exe.getAbsolutePath();
    }

    try
    {
        // Just try invoking java from the system path; this of course
        // assumes "java[.exe]" is /actually/ Java
        final String NAKED_JAVA = "java";
        new ProcessBuilder(NAKED_JAVA).start();

        return NAKED_JAVA;
    }
    catch (IOException e)
    {
        return null;
    }
}

Answer 5

Compilation of All above methods

static String getJavaPath(){

        String tmp1 = System.getProperty("java.home") + "\\bin\\java.exe";
        String tmp2 = System.getProperty("sun.boot.library.path") + "\\java.exe";
        String tmp3 = System.getProperty("java.library.path")+ "\\java.exe";
        if(new File(tmp1).exists()) {
            return tmp1;
        }else if(new File(tmp2).exists()){
            return tmp2;
        }else if(new File(tmp3).exists()) {
            return tmp3;
        }else{
            String[] paths = System.getenv("PATH").split(";");
            for(String path:paths){
                if(new File(path + "\\java.exe").exists()){
                    return path + "\\java.exe";
                }
            }
        }
        return "";
    }

Answer 6

On Windows, the java.library.path System Property begins with the path to the bin directory containing whichever java.exe was used to run your jar file.

This makes sense, because on Windows the first place any executable looks for DLL files is the directory containing the executable itself. So naturally, when the JVM runs, the first place it looks for DLLs is the directory containing java.exe.

You can acquire the path to java.exe as follows:

final String javaLibraryPath = System.getProperty("java.library.path");
final File javaExeFile = new File(
  javaLibraryPath.substring(0, javaLibraryPath.indexOf(';')) + "\\java.exe"
);
final String javaExePath =
  javaExeFile.exists() ? javaExeFile.getAbsolutePath() : "java";

This code is Windows-specific - I hard-coded the path separator (;) and the file separator (). I also put in a fallback to just "java" in case the library path trick somehow doesn't work.

I have tested this with Java 6 and 7 on Windows 7. I tried a 32-bit and 64-bit version of Java.