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 seems like the server is down. Code:' + str(e.code) )
-
1Strange. Can you paste the code you use for opening the url?Manoj Govindan– Manoj Govindan2010-08-12 08:07:09 +00:00Commented Aug 12, 2010 at 8:07
Add a comment
|
2 Answers
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.
1 Comment
zer0c00l
To be 100% safe, you might also want to check for ValueError as urllib2.urlopen will throw ValueError exception when the url entered is invalid or blank.
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
1 Comment
joctee
'URLError' object has no attribute 'code'. It is better to use e.reason.