Run python script in background thread using spring MVC

Question

I have a spring MVC application that need to run a python script that uploads a zip to the api. I need to render the page to the user and run the python script in the background so that the zip gets updates while the user can keep on working in the Spring Application.

    Thread t1 = new Thread(new Runnable() {
                @Override
                public void run() {
                    String[] cmd = {
                            "python",
                            "/Users/rasheenruwisha/final-year-proj/build.py",
                            "ARG 1",
                            "ARG 2",
                    };
                    try {
                        Runtime.getRuntime().exec(cmd);
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }
            });
            t1.start();
            return modelAndView;

This is the current approach I am using, I am not getting any errors but the script does not get executed, is there anything I'm doing wrong?

Abdullajon · Accepted Answer · 2020-02-01 09:29:31Z

2

if you use Ubuntu add "sh" and "-c":

 String[] cmd = {
                            "sh",
                            "-c",
                            "python",
                            "/Users/rasheenruwisha/final-year-proj/build.py",
                            "ARG 1",
                            "ARG 2",
                        };
                        try {
                                Runtime.getRuntime().exec(cmd);
                             } catch (IOException e) {
                                e.printStackTrace();
                             }

answered Feb 1, 2020 at 9:29

Abdullajon

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Comments

Adnan · Accepted Answer · 2020-02-01 18:50:18Z

1

1.You can also use org.python.util.PythonInterpreter

    PythonInterpreter interpreter = new PythonInterpreter();
    try {
     interpreter.execfile("/Users/rasheenruwisha/final-year-proj/build.py","ARG 1,"ARG 
     2");
    } catch (Exception e) {
        e.printStackTrace();
    }

answered Feb 1, 2020 at 18:50

Adnan

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