How do I catch a specific HTTP error in Python?

Question

I have

import urllib2
try:
   urllib2.urlopen("some url")
except urllib2.HTTPError:
   <whatever>

but what I end up is catching any kind of HTTP error. I want to catch only if the specified webpage doesn’t exist (404?).

Asked By: Arnab Sen Gupta

||

Source

Answer 1

Python 3

from urllib.error import HTTPError

Python 2

from urllib2 import HTTPError

Just catch HTTPError, handle it, and if it’s not Error 404, simply use raise to re-raise the exception.

See the Python tutorial.

Here is a complete example for Python 2:

import urllib2
from urllib2 import HTTPError
try:
   urllib2.urlopen("some url")
except HTTPError as err:
   if err.code == 404:
       <whatever>
   else:
       raise

Answered By: Tim Pietzcker

Answer 2

For Python 3.x

import urllib.request
import urllib.error
try:
    urllib.request.urlretrieve(url, fullpath)
except urllib.error.HTTPError as err:
    print(err.code)

Answered By: Lazik

Answer 3

Tim’s answer seems to me as misleading especially when urllib2 does not return the expected code. For example, this error will be fatal (believe or not – it is not uncommon one when downloading urls):

AttributeError: ‘URLError’ object has no attribute ‘code’

Fast, but maybe not the best solution would be code using nested try/except block:

import urllib2
try:
    urllib2.urlopen("some url")
except urllib2.HTTPError as err:
    try:
        if err.code == 404:
            # Handle the error
        else:
            raise
    except:
        ...

More information to the topic of nested try/except blocks Are nested try/except blocks in python a good programming practice?

Answered By: sonavolob

Answer 4

If from urllib.error import HTTPError doesn’t work, try using from requests.exceptions import HTTPError.

Sample:

from requests.exceptions import HTTPError

try:
    <access some url>
except HTTPError:
    # Handle the error like ususal

Answered By: Jubin Ben

How do I catch a specific HTTP error in Python?

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