urllib2ソケットタイムアウトを処理するには？

Question

以下は、タイムアウトしてループ内の次のリンクに続いた他のリンクでも機能しています。しかし、このリンクの私はエラーが発生しました。私はそれがなぜ、どのようにしてそれが起こったときに次のイメージをブラウズするように修正するのか分からない。

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 

エラーは次のとおりです。

[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 <module> 
    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 

Answer 1

明示的にタイムアウト例外をキャッチ：https://docs.python.org/3/library/socket.html#socket.timeout

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()) 

OP： ありがとう！ 私はあなたの提案にこれらの感謝を持っていたし、私の問題は、Python2.7のために解決された：

except socket.timeout as e: 
    print(e) 
    continue 
except socket.error as e: 
    print(e) 
    continue