Question

如果urllib.request.urlopen不是200，我如何避免response.status_code的例外情况？现在它根据请求状态提升URLError或HTTPError。

有没有其他方法可以使用python3基本库进行请求？

如果status_code != 200？

，我如何获得回复标题？

Answer 1

使用try except，代码如下：

from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError
req = Request("http://www.111cn.net /")
try:
    response = urlopen(req)
except HTTPError as e:
    # do something
    print('Error code: ', e.code)
except URLError as e:
    # do something
    print('Reason: ', e.reason)
else:
    # do something
    print('good!')

Answer 2

我从py3 docs找到了一个解决方案

>>> import http.client
>>> conn = http.client.HTTPConnection("www.python.org")
>>> # Example of an invalid request
>>> conn.request("GET", "/parrot.spam")
>>> r2 = conn.getresponse()
>>> print(r2.status, r2.reason)
404 Not Found
>>> data2 = r2.read()
>>> conn.close()

https://docs.python.org/3/library/http.client.html#examples

Answer 3

The docs指出，异常类型HTTPError也可以视为HTTPResponse。因此，您可以从错误响应中获取响应主体，如下所示：

import urllib.request
import urllib.error

def open_url(request):
  try:
    return urllib.request.urlopen(request)
  except urllib.error.HTTPError as e:
    # "e" can be treated as a http.client.HTTPResponse object
    return e

，然后使用如下：

result = open_url('http://www.stackoverflow.com/404-file-not-found')
print(result.status)           # prints 404
print(result.read())           # prints page contents
print(result.headers.items())  # lists headers

Python 3 urllib.request.urlopen

3 个答案: