CODEWITHSUNDEEP
javaprocessbuilder
Chris Snow
Chris Snow

Reputation: 24596

platform independent way to get current java executable path and filename

I need to execute a java application using ProcessBuilder. I also need to use the same java executable in the called process that is being used by the calling application.

Is there a platform independent way to retrieve the current java executable path and file name?

I've seen some code snippets similar to this:

public static boolean isWindows() {
    return (System.getProperty("os.name").toLowerCase().indexOf("win") >= 0);
}

public static String getJavaExecutablePath() {

    String executable = "java";
    if (isWindows()) {
       executable = executable + ".exe";
    }
    File path = new File(System.getProperty("java.home"), "/bin/" + executable);
    return path.getAbsolutePath();
}

The above code will probably need to be improved to use File.separator.

Is the above all that is required, or is there something else to consider, for example case sensitivity?

Ideally, there would be a library available for getting the OS path, but that is probably a question for a different forum.

Upvotes: 0

Views: 622

Answers (1)

Doug Hou
Doug Hou

Reputation: 560

First, the most important thing for your requirement is System.getProperty("java.home"), you've got it already.

Secondly, don't worry about case sensitivity, the path '$JAVA_HOME/bin/java' always be in lower case in every java release, and there are already many other projects depend on this.

Last thing, you don't have to use a '/' or '\' to build the absolute path, consider this:

File path = new File(new File(System.getProperty("java.home"), "bin"), executable);

As you mentioned the File.separator did similar thing, but that is talking about a string, and File talks about file.

Here is a discussion about File path

Upvotes: 1

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