📜  python 从 url 获取响应 - Python 代码示例

📅  最后修改于: 2022-03-11 14:47:04.691000             🧑  作者: Mango

代码示例1
import requests
r = requests.get("https://google.com")
print(r.status_code)
#100- 199 Informational
#200-299 Succes
#300-399 Redirection
#400-499 CLIENT ERROR 
#500-599 SERVER ERROR