Python urllib2 URLError HTTP status code.

Question

I want to grab the HTTP status code once it raises a URLError exception:

I tried this but didn\'t help:

except URLError, e:
    logger.warning( \'It see         


        

Answer 1

You shouldn't check for a status code after catching URLError, since that exception can be raised in situations where there's no HTTP status code available, for example when you're getting connection refused errors.

Use HTTPError to check for HTTP specific errors, and then use URLError to check for other problems:

try:
    urllib2.urlopen(url)
except urllib2.HTTPError, e:
    print e.code
except urllib2.URLError, e:
    print e.args

Of course, you'll probably want to do something more clever than just printing the error codes, but you get the idea.

Answer 2

Not sure why you are getting this error. If you are using urllib2 this should help:

import urllib2
from urllib2 import URLError

try:
    urllib2.urlopen(url)
except URLError, e:
    print e.code