catch specific HTTP error in python

问题:

I want to catch a specific http error and not any one of the entire family.. what I was trying to do is --

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

but what I end up is catching any kind of http error, but I want to catch only if the specified webpage doesn't exist!! probably that's HTTP error 404..but I don't know how to specify that catch only error 404 and let the system run the default handler for other events..ny suggestions??

回答1:

Just catch urllib2.HTTPError, handle it, and if it's not Error 404, simply use raise to re-raise the exception.

See the Python tutorial.

So you could do:

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

回答2:

For Python 3.x

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

回答3:

Tims answer seems to me as misleading. Especially when urllib2 does not return 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, 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?