node.js：如何在前台生成分离的子节点并退出

for child_process.spawn,我希望能够在前台运行子进程并允许节点进程本身退出,如下所示：

切换-exec.js：

'use strict';

var spawn = require('child_process').spawn;

// this console.log before the spawn seems to cause
// the child to exit immediately, but putting it
// afterwards seems to not affect it.
//console.log('hello');

var child = spawn(
  'ping'
, [ '-c', '3', 'google.com' ]
, { detached: true, stdio: 'inherit' }
);

child.unref();

它只是退出而没有任何消息或错误,而不是看到ping命令的输出.

node handoff-exec.js
hello
echo $?
0

所以……在node.js(或根本没有)可以在父节点退出时在前台运行子节点吗？

更新：我发现删除console.log(‘hello’);允许孩子跑,但是,它仍然没有将前台标准控制传递给孩子.

你错过了

// Listen for any response:
child.stdout.on('data', function (data) {
    console.log(data.toString());
});

// Listen for any errors:
child.stderr.on('data', function (data) {
    console.log(data.toString());
});

你不需要child.unref();