如何在node.js中清除child_process.spawn（）上的子进程

Question

The following code:

#!/usr/bin/env node
"use strict";

var child_process = require('child_process');

var x = child_process.spawn('sleep', [100],);

throw new Error("failure");

spawns a child process and exits without waiting the child process exiting.

How can I wait it? I'd like to call waitpid(2) but child_process seems to have no waitpid(2).

ADDED:

Sorry, what I really want is to kill the child process when the parent exists, instead of wait it.

Answer 1

#!/usr/bin/env node
"use strict";

var child_process = require('child_process');

var x = child_process.spawn('sleep', [10]);

x.on('exit', function () {
    throw (new Error("failure"));
});

EDIT:

You can listen for the main process by adding a listener to the main process like process.on('exit', function () { x.kill() })

But throwing an error like this is a problem, you better close the process by process.exit()

#!/usr/bin/env node
"use strict";

var child_process = require('child_process');

var x = child_process.spawn('sleep', [100]);

process.on('exit', function () {
    x.kill();
});

process.exit(1);

Answer 2

#!/usr/bin/env node
"use strict";

var child_process = require('child_process');

var x = child_process.spawn('sleep', [10]);

process.on('exit', function() {
  if (x) {
    x.kill();
  }
});