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Re: exception due to NoneTypeasit a écrit :
> In my program I want to catch exception which is caused by accessing > NoneType object. > > Can anyone suggest me how this can be done ?? Not without the minimal working code exposing your problem, or the full traceback you got. Merely "accessing NoneType object" doesn't by itself raise any exception... I suspect you get the None object where you expected something else and try to access an attribute of this 'something else', and ends up getting an AttributeError, but there are other possible scenarii that might fit your (very poor) description of the problem, so no way too help you without more informations. As a general rule, remember that the traceback is actually meant to *help* finding out what went wring. -- http://mail.python.org/mailman/listinfo/python-list |
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exception due to NoneTypeIn my program I want to catch exception which is caused by accessing
NoneType object. Can anyone suggest me how this can be done ?? -- http://mail.python.org/mailman/listinfo/python-list |
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Re: exception due to NoneTypeOn Nov 7, 10:36 pm, Bruno Desthuilliers
<bdesth.quelquech...@...> wrote: > asit a écrit : > > > In my program I want to catch exception which is caused by accessing > > NoneType object. > > > Can anyone suggest me how this can be done ?? > > Not without the minimal working code exposing your problem, or the full > traceback you got. Merely "accessing NoneType object" doesn't by itself > raise any exception... I suspect you get the None object where you > expected something else and try to access an attribute of this > 'something else', and ends up getting an AttributeError, but there are > other possible scenarii that might fit your (very poor) description of > the problem, so no way too help you without more informations. As a > general rule, remember that the traceback is actually meant to *help* > finding out what went wring. I could have described the error, but the problem is that it's dependent of a third party library.. Let me write the code here... import twitter api = twitter.Api('asitdhal','swordfish') users = api.GetFriends() for s in users: print "##########################################" try: print "user id : " + str(s.id) print "user name : " + s.name print "user location : " + s.location print "user description : " + s.description print "user profile image url : " + s.profile_image_url print "user url : " + s.url print "user status : " + str(s.status) except TypeError: pass look at the except TypeError. This is supposed to catch only exception thrown by NoneType. please help me. -- http://mail.python.org/mailman/listinfo/python-list |
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Re: exception due to NoneTypeasit a écrit :
> On Nov 7, 10:36 pm, Bruno Desthuilliers > <bdesth.quelquech...@...> wrote: >> asit a écrit : >> >>> In my program I want to catch exception which is caused by accessing >>> NoneType object. >>> Can anyone suggest me how this can be done ?? >> Not without the minimal working code exposing your problem, or the full >> traceback you got. Merely "accessing NoneType object" doesn't by itself >> raise any exception... I suspect you get the None object where you >> expected something else and try to access an attribute of this >> 'something else', and ends up getting an AttributeError, but there are >> other possible scenarii that might fit your (very poor) description of >> the problem, so no way too help you without more informations. As a >> general rule, remember that the traceback is actually meant to *help* >> finding out what went wring. > > I could have described the error, but the problem is that it's > dependent of a third party library.. This more often than not translates to "dependent of a wrong use of a 3rd part library". > Let me write the code here... > > import twitter > > api = twitter.Api('asitdhal','swordfish') I hope this is not your actual login. > users = api.GetFriends() > for s in users: > print "##########################################" > try: > print "user id : " + str(s.id) Please learn to use print and string formatting. Here are two ways to get rid of the need to use str(): 1/ print "user id : %s" % s.id 2/ print "user id : ", s.id > print "user name : " + s.name > print "user location : " + s.location > print "user description : " + s.description > print "user profile image url : " + s.profile_image_url > print "user url : " + s.url > print "user status : " + str(s.status) > except TypeError: > pass Silently passing exception is 99.999 out of 100 a very stupid thing to do. > > look at the except TypeError. Yes, I've seen it. It's stupid, and actually make debugging harder. Any statement in this try block could raise a TypeError if the looked up attribute of 's' is not a string. Python is NOT php, and will not happily adds apples and parrots - you can only concatenate strings with strings, not with ints or None or whatever... > This is supposed to catch only exception > thrown by NoneType. Please get your facts straight. 1/ The NoneType doesn't "throw" (the appropriate python term is "raise") any exception by itself 2/ this except clause will catch any TypeError. Try this: try: x = "aaa" + 42 except TypeError, e: print "This has nothing to do with NoneType" > please help me. Help yourself and read the FineManual. Then make appropriate use of string formatting (your best friend for simple formatted outputs) instead of useless string concatenations. -- http://mail.python.org/mailman/listinfo/python-list |
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Re: exception due to NoneType> api = twitter.Api('asitdhal','swordfish')
You just gave the world the account-information to your twitter-account. You'd rather change these asap, or somebody hijacks your account... Diez -- http://mail.python.org/mailman/listinfo/python-list |
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Re: exception due to NoneTypeasit <lipun4u@...> writes:
> for s in users: > print "##########################################" > try: > print "user id : " + str(s.id) > print "user name : " + s.name > print "user location : " + s.location > print "user description : " + s.description > print "user profile image url : " + s.profile_image_url > print "user url : " + s.url > print "user status : " + str(s.status) > except TypeError: > pass Why are you catching TypeError, only to discard it? Ignoring an error doesn't make it go away. The above code should rather use string formatting to interpolate the values you want into a string for output:: import textwrap for user in users: print textwrap.dedent(""" ########################################## user id: %(id)s user name: %(name)s user location: %(location)s user description: %(description)s user profile image url: %(profile_image_url)s user url: %(url)s user status: %(status)s """) % vars(user) -- \ “My classmates would copulate with anything that moved, but I | `\ never saw any reason to limit myself.” —Emo Philips | _o__) | Ben Finney -- http://mail.python.org/mailman/listinfo/python-list |
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