如何检查给定URL中是否存在service-worker？

Question

我可以区分给定的外部网站是否有"服务工作者"？

我最好的猜测是:

1. 从给定的URL检索所有js文件
2. 搜索字符串'sw.js'

(我不确定如何做到这一点)

鉴于数据将是这样的:

const sitesToCheck = {
    'www.site1.com',
    'www.site2.com',
    'www.site3.com',
}

Answer 1

如果要在代码中执行此操作,则只能在与要检查的工作程序相同的原点内执行此操作.在同一个源中,您可以使用getRegistration(url)(它为您提供将解析为URL undefined的ServiceWorkerRegistration对象或URL 的对象),或getRegistrations()(它为您提供ServiceWorkerRegistration对象数组的承诺).例如:

navigator.serviceWorker.getRegistrations().then(registrations => {
    console.log(registrations);
});

Answer 2

正如Nicolas所建议的，这项检查可以使用无头 Chrome 来完成。

下面是一个使用Puppeteer的例子：

'use strict'

const puppeteer = require('puppeteer');

(async() => {

  try {
    // Step 1: launch browser and open a new page.
    const browser = await puppeteer.launch()
    const page = await browser.newPage()

    // Step 2: Go to a URL and wait for a service worker to register.
    var url = 'https://web.dev/'
    await page.goto(url)
    const swTarget = await browser.waitForTarget(target => target.type() === 'service_worker')

    // Step 3a: If a service worker is registered, print URL of SW file to the console 
    if(swTarget) {
      console.log(swTarget._targetInfo['url'])
    }
    // Step 4: Done. Close.
    await browser.close()

  } catch (err) {
    // The process will timeout after 30s, if no service worker is registered
    console.error(err.message)
  }
})()