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Python 获取 HTTP 请求的状态码(200,404 等),不访问整个页面源码，那样太浪费资源：
输入：www.v2ex.com 输出：200 输入：www.v2ex.com/nonexistant 输出：404  | 1 revotu 2017-06-28 09:34:21 +08:00 Python 实用脚本清单 : http://www.revotu.com/python-practical-script-list.html http 不只有 get 方法（请求头部+正文），还有 head 方法，只请求头部。 import httplib def get_status_code(host, path="/"): """ This function retreives the status code of a website by requesting HEAD data from the host. This means that it only requests the headers. If the host cannot be reached or something else goes wrong, it returns None instead. """ try: cOnn= httplib.HTTPConnection(host) conn.request("HEAD", path) return conn.getresponse().status except StandardError: return None print get_status_code("www.v2ex.com") # prints 200 print get_status_code("www.v2ex.com", "/nonexistant") # prints 404 |
 | 2 bolide2005 2017-06-28 09:35:43 +08:00 >>> import requests >>> respOnse= requests.get("http://www.baidu.com") >>> print response.status_code 200 |
 | 3 dd99iii 2017-06-30 08:56:05 +08:00 via iPhone def get_status_code(url): r = requests.head(url) return r.status_code |
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