在 Python 中以交互模式运行 playwright

Question

我正在使用 playwright 使用 Python 来抓取页面。我知道如何使用脚本执行相同的操作，但我正在以交互模式尝试此操作。

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)

我尝试在交互模式下执行此操作：

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

但这给了我一个错误：

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)

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，您也可以使用该wait_for_timeout函数（仅适用于Page对象）并将超时设置为较高值：

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)