Question

I am trying to execute a Bash Shell script from Java and it runs fine using this piece of code.

public void executeScript() {
    try {
        new ProcessBuilder("myscript.sh").start();
        System.out.println("Script executed successfully");
    } catch (IOException e) {
        e.printStackTrace();
    }
}

The above code runs fine asynchronously. But what I would like to achieve is to execute the code synchronously. I would like the Java process to wait until the execution of the script is completed and then execute the next batch of code.

To summarize, I would like the "Print statement - Script executed successfully" to be executed after the batch file ("myscript.sh") completes execution.

Thanks

Was it helpful?

Solution

You want to wait for the Process to finish, that is waitFor() like this

public void executeScript() {
  try {
    ProcessBuilder pb = new ProcessBuilder(
      "myscript.sh");
    Process p = pb.start();     // Start the process.
    p.waitFor();                // Wait for the process to finish.
    System.out.println("Script executed successfully");
  } catch (Exception e) {
    e.printStackTrace();
  }
}

OTHER TIPS

The above code doesn't work if I wanted to move a file from one location to another, so I've fixed it with below code.

class Shell

{

  public static void main(String[] args) {
    try {
      ProcessBuilder pb = new ProcessBuilder("/home/sam/myscript.sh");
      Process p = pb.start();     // Start the process.
      p.waitFor();                // Wait for the process to finish.
      System.out.println("Script executed successfully");
    } catch (Exception e) {
      e.printStackTrace();
      }

  }
}


myscript.sh
#!/bin/bash   
mv -f /home/sam/Download/cv.pdf /home/sam/Desktop/

Use Process#waitFor() to pause the Java code until the script terminates. As in

    try {
        Process p = new ProcessBuilder("myscript.sh").start();
        int rc = p.waitFor();
        System.out.printf("Script executed with exit code %d\n", rc);
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