javascript - firebase 功能中的 Puppeteer 无法启动 chrome

Question

我尝试在 firebase 函数中运行 puppeteer。据我了解 https://github.com/GoogleChrome/puppeteer/blob/master/docs/troubleshooting.md#running-puppeteer-on-google-cloud-functions 它应该可以添加

 "engines": {
    "node": "8"
  },

在 package.json 中。我得到了使用 firebase 部署的积极反馈

 functions: updating Node.js 8 function helloWorld(us-central1)...

不幸的是它兑现了

Error: Error: Failed to launch chrome!
[0408/080847.149912:ERROR:zygote_host_impl_linux.cc(89)] Running as root without --no-sandbox is not supported. See https://crbug.com/638180.


TROUBLESHOOTING: https://github.com/GoogleChrome/puppeteer/blob/master/docs/troubleshooting.md

    at startPuppeteer.then.then.then.catch.error (/srv/index.js:42:15)
    at <anonymous>
    at process._tickDomainCallback (internal/process/next_tick.js:229:7)

在本地测试了相同的代码，该代码有效。

云函数代码失败了。

const functions = require('firebase-functions');

const puppeteer = require('puppeteer');
let browser = ""
let page = ""

const startPuppeteer = async () => {
    browser = await puppeteer.launch();
    page = await browser.newPage()
}

const usePageInPuppeteer = async (url) => {
    await page.goto(url);
    return await page.title()
}

const closePuppeteer = async () => {
    return await browser.close();
}

const runtimeOpts = {
    timeoutSeconds: 500,
    memory: '2GB'
  }

exports.helloWorld = functions
.runWith(runtimeOpts)
.https.onRequest((request, response) => {

    //response.send()
    startPuppeteer()
    .then(() => {
        return usePageInPuppeteer('https://www.google.com')
    })
    .then(returnUse => {
        console.log(returnUse)
        return response.send(returnUse)
    })
    .then(() => {
        return closePuppeteer()
    })
    .catch(error => {
        throw new Error(error)
    });

});

本地测试，有效

const puppeteer = require('puppeteer');

let browser = ""
let page = ""

const startPuppeteer = async () => {
    browser = await puppeteer.launch();
    page = await browser.newPage()
}

const usePageInPuppeteer = async (url) => {
    await page.goto(url);
    return await page.title()
}

const closePuppeteer = async () => {
    return await browser.close();
}


startPuppeteer()
.then(() => {
    return usePageInPuppeteer('https://www.google.com')
})
.then(returnUse => {
    console.log(returnUse)
    return closePuppeteer()
})
.catch(error => {
    throw new Error(error)
});

Answer 1

const startPuppeteer = async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox', '--disable-setuid-sandbox']});
    page = await browser.newPage()
}

效果很好。感谢https://stackoverflow.com/users/2911633/igor-ilic提示