在Google App Engines中,如何在Python中显示已获取URL的页面的HTML源代码？

import urlfetch
from bs4 import BeautifulSoup

url = "https://example.com"

# 使用 urlfetch 库发送 HTTP 请求
result = urlfetch.fetch(url)

# 检查请求是否成功
if result.status_code == 200:
    # 使用 BeautifulSoup 库解析 HTML 源代码
    soup = BeautifulSoup(result.content, "html.parser")
    # 输出 HTML 源代码
    print(soup.prettify())
else:
    print("请求失败，状态码：", result.status_code)

libraries:
- name: urlfetch
  version: "2.0.2"