How to handle urllib2 socket timeouts?
<p>So the following has worked for other links that have timed out and has continued to the next link in the loop. However for this link I got an error. I am not sure why that is and how to fix it so that when it happens it just browses to the next image.</p>

<pre><code>try:
    image_file = urllib2.urlopen(submission.url, timeout = 5)
    with open('/home/mona/computer_vision/image_retrieval/images/'
              + category + '/'
              + datetime.datetime.now().strftime('%y-%m-%d-%s')
              + submission.url[-5:], 'wb') as output_image:
        output_image.write(image_file.read())
except urllib2.URLError as e:
    print(e)
    continue
</code></pre>

<p>The error is:</p>

<pre class="lang-none prettyprint-override"><code>[LOG] Done Getting http://i.imgur.com/b6fhEkWh.jpg
submission id is: 1skepf
[LOG] Getting url:  http://www.redbubble.com/people/crtjer/works/11181520-bling-giraffe
[LOG] Getting url:  http://www.youtube.com/watch?v=Y7iuOZVJhs0
[LOG] Getting url:  http://imgur.com/8a62PST
[LOG] Getting url:  http://www.youtube.com/watch?v=DFZFiFCsTc8
[LOG] Getting url:  http://i.imgur.com/QPpOFVv.jpg
[LOG] Done Getting http://i.imgur.com/QPpOFVv.jpg
submission id is: 1f3amu
[LOG] Getting url:  http://25.media.tumblr.com/tumblr_lstla7vqK71ql5q9zo1_500.jpg
Traceback (most recent call last):
  File "download.py", line 50, in &lt;module&gt;
    image_file = urllib2.urlopen(submission.url, timeout = 5)
  File "/usr/lib/python2.7/urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 404, in open
    response = self._open(req, data)
  File "/usr/lib/python2.7/urllib2.py", line 422, in _open
    '_open', req)
  File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 1214, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.7/urllib2.py", line 1187, in do_open
    r = h.getresponse(buffering=True)
  File "/usr/lib/python2.7/httplib.py", line 1051, in getresponse
    response.begin()
  File "/usr/lib/python2.7/httplib.py", line 415, in begin
    version, status, reason = self._read_status()
  File "/usr/lib/python2.7/httplib.py", line 371, in _read_status
    line = self.fp.readline(_MAXLINE + 1)
  File "/usr/lib/python2.7/socket.py", line 476, in readline
    data = self._sock.recv(self._rbufsize)
socket.timeout: timed out
</code></pre>

<p>Explicitly catch the timeout exception: <a href="https://docs.python.org/3/library/socket.html#socket.timeout" rel="nofollow">https://docs.python.org/3/library/socket.html#socket.timeout</a></p>

<pre><code>try:
    image_file = urllib2.urlopen(submission.url, timeout = 5)
except urllib2.URLError as e:
    print(e)
    continue
except socket.Timeouterror:
    print("timed out")
    # Your timeout handling code here...
else:
    with open('/home/mona/computer_vision/image_retrieval/images/'+category+'/' + datetime.datetime.now().strftime('%y-%m-%d-%s') + submission.url[-5:], 'wb') as output_image:
        output_image.write(image_file.read())
</code></pre>

<p>OP:
Thanks!
I had these thanks to your suggestion and my problem was solved for Python2.7:</p>

<pre><code>except socket.timeout as e:
    print(e)
    continue
except socket.error as e:
    print(e)
    continue
</code></pre>

python exception urllib2 urllib continue