CODEWITHSUNDEEP
pythonexceptionerror-handling
KKa
KKa

Reputation: 408

URLError in Python

I am trying to understand if I can handle the following error in python.

So I have a program which repeatedly calls the following line:

candidate = urllib2.urlopen(absolute_path)

After running my program for some seconds, I turned off my wifi connection, and got the following error:

File "crawler.py", line 28, in urlQuery
  candidate = urllib2.urlopen(absolute_path)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 127, in urlopen
  return _opener.open(url, data, timeout)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 404, in open
  response = self._open(req, data)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 422, in _open
'_open', req)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 382, in _call_chain
  result = func(*args)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1214, in http_open
  return self.do_open(httplib.HTTPConnection, req)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1184, in do_open
  raise URLError(err)
urllib2.URLError: <urlopen error [Errno 65] No route to host>

Is there any way I can handle this error?

Upvotes: 1

Views: 4064

Answers (1)

sk11
sk11

Reputation: 1824

Depends what you want to do when you get this Exception. You can use try-except, standard Python's exception handling technique.

try:
    candidate = urllib2.urlopen(absolute_path)
#except Exception as e: # catches any exception
except urllib2.URLError as e: # catches urllib2.URLError in e
    print ('WiFi connection perhaps lost !! Trying one more time...')
    try:
        candidate = urllib2.urlopen(absolute_path)
    except:
        print ('WiFi connection really lost !! Bailing out..')
        print (e) # print outs the exception message

Upvotes: 1

Related Questions