node.js:如何在前台生成分离的子节点并退出

Question

根据文档,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:

handoff-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();

ping它只是退出而没有任何消息或错误,而不是看到命令的输出.

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');允许孩子运行,但是,它仍然没有将前台标准输入控制传递给孩子.

Answer 1

解决方案

问题中的代码实际上是正确的。当时节点中存在一个合法的错误。

'use strict';

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

console.log("Node says hello. Let's see what ping has to say...");

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

child.unref();

上面的代码片段将有效地运行，就像它在 shell 的后台运行一样：

ping -c 3 google.com &