如何从 puppeteer/playwright JSelement 节点中删除一个子节点，然后获取 innerText

Question

我可以使用 playwright/puppeteer 获取单元格。我想分别捕获以下两个值 - 日期和状态。

我有以下代码：

  let allCells = await allRows[0].$$('[role="cell"]');

  let ele = await allCells[0].$('.description');
  let status = await (await ele.getProperty("innerText")).jsonValue();
       // I can get the status as 'uploaded' just fine using this

  allCells[0].removeChild(ele);    // this throws an error

  let uploadDate = await (await allCells[0]("innerText")).jsonValue();

它抛出的错误是：类型错误：allCells[0].removeChild 不是函数

console.log( allCells[0] ) returns: JSHandle@....

这是 HTML 的相关部分：

<html>
<body>
  <div role="cell" class="cell-body">
     <!---->Jul 11, 2021
     <div class="description">
        uploaded
     </div>
  </div>
</body>
</html>

Answer 1

遗憾的是，您无法在 JS 或 puppeteer (Node.js) 上下文中调用 web API 方法 (.removeChild) 或元素句柄。

您可以尝试使用类似这样的方法获取浏览器上下文中的所有数据（.childNodes[0] 只会给您第一个文本节点，直到 <div class="description"> 元素）：

import puppeteer from 'puppeteer';

const browser = await puppeteer.launch();

const html = `
  <html>
  <body>
    <div role="cell" class="cell-body">
       Jul 11, 2021
       <div class="description">
          uploaded
       </div>
    </div>
  </body>
  </html>`;

try {
  const [page] = await browser.pages();

  await page.goto(`data:text/html,${html}`);

  const data = await page.evaluate(() => {
    const date = document.querySelector('div.cell-body').childNodes[0].textContent.trim();
    const description = document.querySelector('div.description').innerText;
    return [date, description];
  });
  console.log(data);
} catch (err) { console.error(err); } finally { await browser.close(); }

如何从 puppeteer/playwright JSelement 节点中删除一个子节点，然后获取 innerText

How to remove a child from a puppeteer/playwright JSelement node and then fetch innerText

javascript

htmlelements

puppeteer

playwright