我有一个PHP脚本,我用这段代码通过URL参数调用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()
然而,当我想在访问页面后打印出页面源代码时,我收到了以下错误信息:
Error in playwright script.
因为我尝试过的代码无法正常工作:
print(page.innerHTML("*"))
需要帮忙吗?