How to get status code for urllib2 in python 2.7

Question

How do I get status code in python 2.7 for urllib2? I dont want to use requests. I need urllib2.

    request = urllib2.Request(url, headers=headers)
    contents = urllib2.urlopen(request).read()
    print request.getcode()
    contents = json.loads(contents) 

     <type 'exceptions.AttributeError'>, AttributeError('getcode',), <traceback object at 0x7f6238792b48>

Thanks

[This may help](http://stackoverflow.com/a/1726434/736937) – jedwards Mar 12 '15 at 17:04 — jedwards, Mar 12 '15 at 17:04

score 12 · Accepted Answer · answered Mar 12 '15 at 17:05

12

Just take a step back:

result = urllib2.urlopen(request)
contents = result.read()
print result.getcode()

answered Mar 12 '15 at 17:05

mdurant

21,368
2
33
59

score 6 · Answer 2 · answered Mar 12 '15 at 17:07

6

use getcode()

>>> import urllib
>>> a=urllib.urlopen('http://www.google.com/asdfsf')
>>> a.getcode()
404
>>>

for urllib2

try:
    urllib2.urlopen('http://www.google.com/asdfsf')
except urllib2.HTTPError, e:
    print e.code

will print

answered Mar 12 '15 at 17:07

Nishant Nawarkhede

6,960
9
49
67

This answer assummes that the `urllib` method doesn't throw an exception. But it does. I tried to open a url that returns error code 401 (Unauthorized) and it threw an exception on the `urlopen` call. – frakman1 Aug 15 '20 at 15:29

How to get status code for urllib2 in python 2.7

2 Answers2