How can I kill a child_process spwaned by node.js/electron correctly in windows?

Question

I am using Electron, Python and Flask to develop an app in Windows.

I use

require('child_process').spawn('python', [_script,port]) 

to start a child_process, but I cannot kill this child_process correctly.

When the electron app was closed, there was still a process called 'python' working in the background.

I tried almost every way, for instance p.kill(), or using taskkill.

Here is the python code:

from flask import Flask
import logiccode as mine

app = Flask(__name__)

@app.route('/hello')
def hello_world():
   return mine.HelloWorld()


if __name__ == '__main__':
   app.run(debug = True)

//create
pyProc = require('child_process').spawn('python', [_script,port])
    if (pyProc != null) {
        console.log('child process success\n')
        console.log(pyProc.pid);
    }

//kill

if(process.platform=='win32'){
        require('child_process').spawn('taskkill',['/pid','/f','/t',pyProc.pid]);
    }else{
        pyProc.kill();
        pyProc = null
        pyPort = null
    }

If there is some error in the code, please tell me how I can kill this child_process correctly.

Answer 1

I don't see any reason for the Windows-specific branch there, I assume you added it when pyProc.kill() didn't work.

kill defaults to SIGTERM. I'd expect that to work, but if you're seeing it not work, you can use the nuclear option of SIGKILL instead. As it says in signal(7) :

The signals SIGKILL and SIGSTOP cannot be caught, blocked, or ignored.

So:

//kill
pyProc.kill('SIGKILL');
pyProc = null
pyPort = null

Answer 2

Hope it is not too late but I ran into the same problem, the reason why flask cannot be killed is because debug mode is set to True. Just call app.run() and node will kill flask.

if __name__ == '__main__':
   app.run()