等待所有子进程完成才能继续

Question

我想知道是否可以等待使用 spawn 函数创建的所有子进程完成后再继续执行。

我的代码如下所示：


    const spawn = window.require('child_process').spawn;

    let processes = [];
    let thing = [];
    // paths.length = 2
    paths.forEach((path) => {
        const pythonProcess = spawn("public/savefile.py", ['-d', '-j', '-p', path, tempfile]);

        pythonProcess.on('exit', () => {
            fs.readFile(tempfile, 'utf8', (err, data) => {
                thing.push(...)
            });
        });

        processes.push(pythonProcess);
    });

    console.log(processes) // Here we have 2 child processes
    console.log(thing) // empty array.. the python processes didnt finish yet

    return thing // of course it doesn't work. I want to wait for all the processes  to have finished their callbacks to continue

如你所料，我想知道如何同时获取所有python脚本运行，并等待它们全部完成才能继续我的js代码.

我是运行节点10.15.3

谢谢

Answer 1

ForEach 将 Promise 推入 Promise 和 Promise.all()

的数组

等待所有子进程完成才能继续

Wait for all child processes to finish to continue

asynchronous

child-process

node.js