Python 3 urllib.request.urlopen
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Python 3 urllib.request.urlopen
提问by Bogdan Ruzhitskiy
How can I avoid exceptions from urllib.request.urlopenif response.status_codeis not 200? Now it raise URLErroror HTTPErrorbased on request status.
urllib.request.urlopen如果response.status_code不是 200，如何避免异常？现在它提高URLError或HTTPError基于请求状态。
Is there any other way to make request with python3 basic libs?
有没有其他方法可以使用 python3 基本库发出请求？
How can I get response headers if status_code != 200?
如果 ，我如何获得响应标头status_code != 200？
回答by Bogdan Ruzhitskiy
I found a solution from py3 docs
我从 py3 文档中找到了解决方案
>>> 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()
回答by thinkerou
Use try except, the below code:
使用try except，下面的代码：
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!')
