在 python 中以交互模式运行 playwright

Question

我正在使用剧作家使用 python 来抓取页面。 我知道如何使用脚本来做同样的事情，但我是在交互模式下尝试的。

from playwright.sync_api import Playwright, sync_playwright, expect
import time

def run(playwright: Playwright) -> None:
    browser = playwright.chromium.launch(headless=False)
    context = browser.new_context()
   
    page = context.new_page()
    page.goto("https://www.wikipedia.org/")
    
    context.close()
    browser.close()
with sync_playwright() as playwright:
    run(playwright)

我试图在交互模式下这样做：

>>> from playwright.sync_api import Playwright, sync_playwright, expect
>>> playwright = sync_playwright()
>>> browser = playwright.chromium.launch(headless=False)

但这给了我错误：

Traceback (most recent call last):
  File "C:\Users\hpoddar\AppData\Local\Programs\Python\Python310\lib\idlelib\run.py", line 578, in runcode
    exec(code, self.locals)
  File "<pyshell#2>", line 1, in <module>
AttributeError: 'PlaywrightContextManager' object has no attribute 'chromium'

Answer 1

使用.start()方法：

>>> from playwright.sync_api import Playwright, sync_playwright, expect
>>> playwright = sync_playwright().start()
>>> browser = playwright.chromium.launch(headless=False)
>>> page = browser.new_page()

或者，如果您只想要一个交互式浏览器，而不关心交互式 shell，您也可以使用wait_for_timeout function 代替（仅适用于Page对象）并将超时设置为高值：

from playwright.sync_api import sync_playwright

with sync_playwright() as playwright:
    browser = playwright.chromium.launch(headless=False)
    page = browser.new_page()
    page.wait_for_timeout(10000)