如何使用 playwright-python 处理多个页面？

Question

如何使用playwright-python 收听新页面？

在 JavaScript 中，它将被记录为：

const playwright = require("playwright");

(async () => {
  const browser = await playwright.chromium.launch();
  const context = await browser.newContext();
  const page = await context.newPage();
  context.on("page", async newPage => {
    console.log("newPage", await newPage.title())
  })

  // emulate some opening in a new tab or popup
  await page.evaluate(() => window.open('https://google.com', '_blank'))
  // Keep in mind to have some blocking action there so that the browser won't be closed. In this case we are just waiting 2 seconds.
  await page.waitForTimeout(2000)
  await browser.close();
})();

变成Python

from playwright import sync_playwright

with sync_playwright() as p:
    browser = p.chromium.launch(
        headless=False,
        executablePath='C:/Program Files/Google/Chrome/Application/chrome.exe'
    )
    context = browser.newContext()
    page = context.newPage()

'''
how to do in Python?
  context.on("page", async newPage => {
    console.log("newPage", await newPage.title())
  })

  // emulate some opening in a new tab or popup
  await page.evaluate(() => window.open('https://google.com', '_blank'))
'''

    page.waitForTimeout(2000)
    browser.close()

Answer 1

感谢@hardkoded 这是解决方案：

from playwright import sync_playwright

def newPage(page):
   print("newPage() page title:", page.title())

with sync_playwright() as p:
    browser = p.chromium.launch(
        headless=False,
        executablePath='C:/Program Files/Google/Chrome/Application/chrome.exe'
    )
    context = browser.newContext()
    page = context.newPage()

    context.on("page", lambda page: newPage(page))

    page.evaluate('''() => {
        window.open('https://google.com', '_blank')
    }''')
       
    page.waitForTimeout(2000)
    browser.close()

Answer 2

如果您需要在没有事件监听器的情况下处理新页面（即通过单击链接打开新标签），您可以尝试以下代码：

    from playwright import sync_playwright

    with sync_playwright() as p:
        browser = p.chromium.launch()
        context = browser.newContext()
        page = context.newPage()
        page.goto('<site url>')
        with context.expect_page() as tab:
            page.click('.newTabByLink')
        # do some steps
        ...
        tab.close()
        
        browser.close()