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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