Janzert schrieb:
> 2. Why not something like the following (untested code)?
> @property
> def permissions(self):
> p_q = session.query(Permission).join(['tg_group', 'tg_user'])
> p_q.filter_by(user_id = self.user_id)
> return p_q.all()
I would write something like that in a long statement, without the
arbitrary p_q variable and avoid using hard coded table names.
The other problem is that this code doesn't work at all.
Also, the method must return a set, not a list with duplicates.
There are basically two ways to collect the permissions via SQL.
Either you make the two joins, creating all possible ways a user can
have a certain permission, and then remove the duplicates, like that:
def permissions(self):
return set(Permission.query.join(Permission.groups,
Group.users).filter_by(user_id=self.user_id).all())
Or you create the query using exist clauses, so that the query does not
produce any duplicates in the first place:
def permissions(self):
return set(Permission.query.filter(Permission.groups.any(
Group.users.any(User.user_id==self.user_id))).all())
The problem with this latter solution only works with SA >= 0.4 (and the
former solution even only with SA >= 0.5).
(Also, both solutions do a last join with the user table that is
actually unnecessary. I currently see no elegant way to avoid this. It
does not harm much, though.)
-- Christoph
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