Javascript 如何在不获取其子元素的文本内容的情况下获取元素的文本内容

我不认为剧作家有这样的天赋，所以进入evaluate可能是最好的方法:

const text = await page
  .locator("span", {has: page.locator("time")})
  .evaluate(el => el.firstChild.textContent);

为了将其推广到具有多个文本 node 或父 node 内的任意位置的情况，

const text = await page
  .locator("span", {has: page.locator("time")})
  .evaluate(el =>
    [...el.childNodes]
      .filter(e => e.nodeType === Node.TEXT_NODE)
      .map(e => e.textContent)
  );

预计会根据需要对文本进行裁剪和连接.例如:

const playwright = require("playwright"); // ^1.39.0

const html = `
<p>
  a <b>ignore this</b>
</p>
<p> b <b>ignore this</b> c </p>
<p> d <b>ignore this</b> e </p>`;

let browser;
(async () => {
  browser = await playwright.firefox.launch();
  const page = await browser.newPage();
  await page.setContent(html);
  const text = await page
    .locator("p")
    .evaluateAll(els =>
      els.map(el =>
        [...el.childNodes]
          .filter(
            e =>
              e.nodeType === Node.TEXT_NODE &&
              e.textContent.trim()
          )
          .map(e => e.textContent.trim())
      )
    );
  console.log(text); // => [ [ 'a' ], [ 'b', 'c' ], [ 'd', 'e' ] ]
})()
  .catch(err => console.error(err))
  .finally(() => browser?.close());

作为mentioned in the comments，如果您在测试中断言，最好使用

await expect(locator).toHaveText(/^\s*The time is:/m);