捕获http错误

Question

例如,如何捕获python和urllib(2)中页面的404和403错误？

有没有大型包装的快速方法？

添加信息(堆栈跟踪):

Traceback (most recent call last):
  File "test.py", line 3, in <module>
    page = urllib2.urlopen("http://localhost:4444")
  File "/usr/lib/python2.6/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.6/urllib2.py", line 391, in open
    response = self._open(req, data)
  File "/usr/lib/python2.6/urllib2.py", line 409, in _open
    '_open', req)
  File "/usr/lib/python2.6/urllib2.py", line 369, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.6/urllib2.py", line 1161, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.6/urllib2.py", line 1136, in do_open
    raise URLError(err)
urllib2.URLError: <urlopen error [Errno 111] Connection refused>

Answer 1

import urllib2 
try:
   page = urllib2.urlopen("some url")
except urllib2.HTTPError, err:
   if err.code == 404:
       print "Page not found!"
   elif err.code == 403:
       print "Access denied!"
   else:
       print "Something happened! Error code", err.code
except urllib2.URLError, err:
    print "Some other error happened:", err.reason

在您的情况下,错误发生在可以构建HTTP连接之前 - 因此您需要添加另一个捕获的错误处理程序URLError.但这与404或403错误无关.

Answer 2

req = urllib2.Request('url')
>>> try:
>>>     urllib2.urlopen(req)
>>> except urllib2.URLError, e:
>>>     print e.code
>>>     print e.read()