exception due to NoneType

[Bruno Desthuilliers]

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.

 

[asit]

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

Can anyone suggest me how this can be done ??

 

[asit]

asit a écrit :



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.

 

[Bruno Desthuilliers]

asit a écrit :

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.

 

[Diez B. Roggisch]

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