exception due to NoneType

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Re: exception due to NoneType

by Bruno Desthuilliers :: Rate this Message:

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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.
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exception due to NoneType

by asit dhal :: Rate this Message:

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In my program I want to catch exception which is caused by accessing
NoneType object.

Can anyone suggest me how this can be done ??
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Re: exception due to NoneType

by asit dhal :: Rate this Message:

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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..

Let me write the code here...

import twitter

api = twitter.Api('asitdhal','swordfish')
users = api.GetFriends()
for s in users:
    print
    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.
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Re: exception due to NoneType

by Bruno Desthuilliers :: Rate this Message:

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asit 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
>     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.

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Re: exception due to NoneType

by Diez B. Roggisch-2 :: Rate this Message:

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> 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
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Re: exception due to NoneType

by Ben Finney-10 :: Rate this Message:

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asit <lipun4u@...> writes:

> for s in users:
>     print
>     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)

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Ben Finney
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