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我有 PHP 脚本，我用这个带有 URL 参数的代码调用 python 函数：

import json
import sys
import urllib.parse
link = urllib.parse.unquote(sys.argv[1])
from playwright.sync_api import sync_playwright
with sync_playwright() as p:
    browser = p.chromium.launch()
    context = browser.new_context(user_agent='Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/89.0.4389.114 Safari/537.36')
    page = context.new_page()
    cookie_file = open('./cookies.json')
    cookies = json.load(cookie_file)
    print(cookies)
    context.add_cookies(cookies)
    page.goto(link)
    try:
        page.wait_for_timeout(10000)
        print(page.innerHTML("*"))
        page.close()
        context.close()
        browser.close()      
    except Exception as e:
        print("Error in playwright script.")
        page.close()
        context.close()
        browser.close()

但是，当我在访问页面后想打印页面源时，我收到

Error in playwright script.

因为我尝试的代码不起作用：

print(page.innerHTML("*"))

有什么帮助吗？

4

1 回答 1

2

要获取页面的完整 HTML 内容，您可以使用page.content().

于 2021-04-22T11:27:58.687 回答