如何在Node.js中等待子进程完成? [英] How to wait for a child process to finish in Node.js?
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问题描述
我正在通过Node.js中的子进程运行Python脚本,如下所示:
I'm running a Python script through a child process in Node.js, like this:
require('child_process').exec('python celulas.py', function (error, stdout, stderr) {
child.stdout.pipe(process.stdout);
});
,但是Node不等待它完成。我该如何等待进程完成?
but Node doesn't wait for it to finish. How can I wait for the process to finish?
编辑:是否可以通过在从主脚本调用的模块中运行子进程来做到这一点?
Is it possible to do this by running the child process in a module I call from the main script?
推荐答案
您应该使用exec-sync
You should use exec-sync
这可以让脚本等待您执行完了
That allow your script to wait that you exec is done
真正易于使用:
var execSync = require('exec-sync');
var user = execSync('python celulas.py');
看看:
https://www.npmjs.org/package/exec-sync
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