Puppeteer:点击输入标签按钮后不刷新的页面抓取html

问题描述 投票:0回答:1

我在点击输入标签按钮后尝试抓取一些html。我点击了带有page.evaluate()的按钮,因为page.click()似乎不适用于输入标签按钮。我尝试使用headless进行可视化调试:在puppeteer启动选项中使用false来验证浏览器是否确实导航到点击按钮后的点。我不确定为什么page.content()在单击按钮之前返回html而不是事件发生后的html。

const puppeteer = require('puppeteer');
const url = 'http://www.yvr.ca/en/passengers/flights/departing-flights';
const fs = require('fs');
const tomorrowSelector = '#flights-toggle-tomorrow'

puppeteer.launch().then(async browser => {
    const page = await browser.newPage();
    await page.goto(url);
    await page.evaluate((selector)=>document.querySelector(selector).click(),tomorrowSelector);
    let html = await page.content();

    await fs.writeFile('index.html', html, function(err){
        if (err) console.log(err);
        console.log("Successfully Written to File.");
    });
   await browser.close();
  });
javascript node.js web-scraping puppeteer
1个回答
2
投票

您可以单击收音机的标签。此外,您需要等待某些已更改状态的标志(对于XHR / fetch响应或新选择器)。例如,此代码适用于我,但您可以使用任何其他条件或只是等待几秒钟。

const fs = require('fs');
const puppeteer = require('puppeteer');

const url = 'http://www.yvr.ca/en/passengers/flights/departing-flights';

const tomorrowLabelSelector = 'label[for=flights-toggle-tomorrow]';
const tomorrowLabelSelectorChecked = '.yvr-form__toggle:checked + label[for=flights-toggle-tomorrow]';

puppeteer.launch({ headless: false }).then(async (browser) => {
  const page = await browser.newPage();
  await page.goto(url);
  await Promise.all([
    page.click(tomorrowLabelSelector),
    page.waitForSelector(tomorrowLabelSelectorChecked),
  ]);

  const html = await page.content();

  await fs.writeFile('index.html', html, (err) => {
    if (err) console.log(err);
    console.log('Successfully Written to File.');
  });
  // await browser.close();
});
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