Typo: FriendShip(key= not FriendShip(parent=
On Tuesday, November 20, 2012 2:48:25 PM UTC-6, Jesse wrote:
>
> Don't model it like this :).
>
> Seriously, do something like this:
>
> class FriendShip(db.Model):
> account_id = db.IntegerProperty() # Sure use refprop if you want,
> waste of space IMHO
>
> Then create it like this:
>
> friendship = FiendShip(parent=db.Key.from_path('User', USER_1_ID,
> 'Friend', USER_2_ID), account_id=USER_2_ID)
> # Then make another one under the other guy:
> friendship_sym = FiendShip(parent=db.Key.from_path('User', USER_2_ID,
> 'Friend', USER_1_ID), account_id=USER_1_ID)
>
> db.put([friendship, friendship_sym])
>
> That does this do?
>
> * Testing for friendship: db.get(db.Key.from_path('User', USER_1_ID,
> 'Friend', USER_2_ID)) -- if not None they are friends.
> * Friends of user one? FriendShip.all(ancestor=db.Key.from_path('User',
> USER_1_ID)) # Note--this is always consistent! Woohoo!
> * Everyone that is friends with user 2?
> FriendShip.all().filter('account_id', USER_2_ID) # Not always consistent,
> but pretty close :)
>
> -Jesse
>
>
> On Tuesday, November 20, 2012 12:45:34 PM UTC-6, Emmanuel Mayssat wrote:
>>
>> I have a model as follow
>>
>> class User(db.Model):
>> name = db.StringProperty(required=True)
>>
>> class Friendship(db.Model):
>> user1 = db.ReferenceProperty(required=True,
>> collection_name="user1_set")
>> user2 = db.ReferenceProperty(required=True,
>> collection_name="user2_set")
>>
>> I can create users and friendship relationship between them.
>> Now the question is:
>> How can I find if to people are already friends?
>>
>
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