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How to Execute a Java Application Independently of Its Location?

Nov 06, 2024 am 10:36 AM

How to Execute a Java Application Independently of Its Location?

Executing a Java Application Separately

One may wish to execute a Java application independently of its location using its name in a way that is not dependent on the operating system.

Implementation

The Java system properties provide data that enables the development of the path to the java command and the classpath in a platform-independent manner.

The following code demonstrates this implementation:

<code class="java">public final class JavaProcess {
    public static int exec(Class klass, List<String> args) throws IOException, InterruptedException {
        String javaHome = System.getProperty("java.home");
        String javaBin = javaHome + File.separator + "bin" + File.separator + "java";
        String classpath = System.getProperty("java.class.path");
        String className = klass.getName();

        List<String> command = new LinkedList<>();
        command.add(javaBin);
        command.add("-cp");
        command.add(classpath);
        command.add(className);
        command.addAll(args);

        ProcessBuilder builder = new ProcessBuilder(command);
        Process process = builder.inheritIO().start();
        process.waitFor();
        return process.exitValue();
    }
}</code>
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You can execute this method as follows:

<code class="java">int status = JavaProcess.exec(MyClass.class, args);</code>
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Note that this method accepts the actual class instead of its name as a string because it must be included in the classpath.

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