木偶戏如何处理异步任务？

Question

我是nodejs和puppeteer的新手，这对我来说是个问题：

我想处理多页异步，有办法吗？

我的解决方案

{
    path: "/project/:name",
    name: "projectDetails",
    props: true,
    component: () =>
      import(/* webpackChunkName: "ProjectDetails"*/ "../views/ProjectDetails")
},

问题

// main.js

(async () => {
    ...
    await Promise.all([
        twitter.Post(text, uploadImages),
        weibo.Post(text, uploadImages),
        zhihu.Post(text, uploadImages)
    ])
    ...
})();


// twitter/weibo/zhihu.js
async Post() {
    const browser = await puppeteer.launch({ headless: false, userDataDir: './browser' });
    const page = await browser.newPage();
}

问题在于木偶只能限制2个进程，而在限制第3个进程时会捕获错误。

Answer 1

const params = {
  args: [
    '--incognito',
    '--ignore-certificate-errors',
    '--no-sandbox',
    '--disable-setuid-sandbox',
    '--window-size=1920,1080',
    '--disable-accelerated-2d-canvas',
    '--disable-gpu'

    // '--unlimited-storage',
    // '--no-startup-window',
    // '--disable-dev-shm-usage',
    // '--disable-crash-reporter',
    // '--disable-breakpad'
  ],
  headless: false
};

(async () => {
    ...
    let promiseArray=[];

    promiseArray.push(twitter.Post(text, uploadImages));
    promiseArray.push(weibo.Post(text, uploadImages));
    promiseArray.push(zhihu.Post(text, uploadImages));

    await Promise.all(promiseArray).then(()=>{
       //success
    }).catch((err)=>console.log(err));

    ...
})();


async function Post() {
  return puppeteer
    .launch(params)
    .then(async (browser) => {
       const page = await browser.newPage();
    });
}