node.js - 从 Web 访问 Telegram 频道时如何关闭弹出窗口？

Question

例如，我访问了 https://t.me/movocashtransfers。将出现一个弹出窗口，询问您是否要打开 Telegram 应用程序。我想关闭它，因为我必须关闭它才能抓取网络信息。

const puppeteer = require('puppeteer')

async function test () {
  try {
    let browser = await puppeteer.launch({
      headless: false,
    })
    let page = await browser.newPage()
    let override = Object.assign(page.viewport(), { width: 1920, height: 1080 })
    await page.setViewport(override)

    page.on('dialog', async dialog => {
      console.log(dialog.message())
      await dialog.dismiss()
      let channelMembers = document.querySelectorAll('.tgme_page.tgme_page_post .tgme_page_extra')[0].innerHTML

      console.log('channelMembers:', channelMembers)
    })

    await page.goto(`https://t.me/movocashtransfers`, { waitUntil: 'networkidle0' })

  } catch (e) {
    console.log('e:', e)
  }
}

test()

这是弹出窗口的样子：

标签： node.jsweb-crawlerpuppeteer