使用 Playwright 在 Chrome 中获取标签名称

Question

我正在尝试使用 playwright 获取选项卡名称，但我的代码仅输出页面 url。

例如，我想从 go 到https://www.rottentomatoes.com/并打印选项卡上显示的名称：

烂番茄：电影/电视节目/电影预告片...

我尝试使用 page.title 和其他一些选项，但它不起作用。

from playwright.sync_api import sync_playwright

website = "https://www.rottentomatoes.com/"


p = sync_playwright().start()
browser = p.chromium.launch(headless=False)
page = browser.new_page()
page.goto(website)
print(page.content)
print(page.inner_text)
print(page.title)
print(page.context)

我的 output 是：

<bound method Page.content of <Page url='https://www.rottentomatoes.com/'>>
<bound method Page.inner_text of <Page url='https://www.rottentomatoes.com/'>>
<bound method Page.title of <Page url='https://www.rottentomatoes.com/'>>

Answer 1

正如其他用户在评论中提到的那样。 您需要在方法inner_text(Selector)选择器作为参数传递以获取值。

from playwright.sync_api import sync_playwright

website = "https://www.rottentomatoes.com/"


p = sync_playwright().start()
browser = p.chromium.launch(headless=False)
page = browser.new_page()
page.goto(website)
print(page.inner_text("rt-header-nav[slot='nav-dropdowns'] [slot='movies']  a[slot='link']"))
print(page.inner_text("rt-header-nav[slot='nav-dropdowns'] [slot='tv']  a[slot='link']"))
print(page.inner_text("rt-header-nav[slot='nav-dropdowns'] [slot='trivia']  a[slot='link']"))

这将打印在控制台中。

MOVIES
TV SHOWS
MOVIE TRIVIA