Thanks for the help, Karen, Solved :D
Django is really so simple... I just keep thinking it has limits... The
limit is only the mind.
--------------------------
Karen's way:
*User.objects.filter( profile__paid='N' )* #works great!
Or even another way I found now:
*profiles= UserProfile.objects.filter(paid='N')*
and in template, to access the User object just call:
(...)
{% *for profile in profiles*%}
<p>User {% *profile.user.username* %} haven't payed yet</p>
(...)
By this way I just rendered the profile and accessed the super class
Thanks Karen.
2009/11/14 Karen Tracey <kmtra...@gmail.com>
> On Sat, Nov 14, 2009 at 11:21 AM, Carlos Ricardo Santos <
> carlosricardosan...@gmail.com> wrote:
>
>> Hi,
>>
>> I extended the user class with a UserProfile, with the Django docs
>> tutorial.
>> But now I need to do a query to list all users that haven't payed
>> (pay='No').
>>
>>
>> I tried something like this: *User.objects.filter(
>> User.get_profile().paid='No' )*
>> or *User.objects.filter( User.profile.paid='No' )*
>>
>>
> The user profile is just a model with a ForeignKey to User, so you can use
> standard ORM methods for accessing it from the User model, you do not have
> to go through get_profile(). So try:
>
> User.objects.filter(userprofile__paid='No')
>
> Karen
>
>
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Cumprimentos,
Carlos Ricardo Santos
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