Person
has one Building
.
Person
has many Group
I want to return all of the people
from a certain building
who do not have any Group
in their groups
collection.
Maybe I can search by people who have a group list that has a length of 0? Something like:
unassigned=Person.query.filter(Person.building==g.current_building,Person.groups.any()).all()
Use negation ( ~
) with any
:
q = session.query(Person)
q = q.filter(Person.building == g.current_building)
q = q.filter(~Person.groups.any())
any
is more powerful than needed in your case, but it will do the job just fine.
First count the groups per building, then filter on that count.
gc = session.query(
Person.id,
db.func.count(Group.id).label('gc')
).join(Person.groups).group_by(Person.id).subquery()
unassigned = session.query(Person).join(
(gc, gc.c.person_id == Person.id)
).filter(
Person.building == g.current_building,
gc.c.gc == 0
).all()
If you need to know if a relation have records after the query use the count()
method on the relation:
persons = session.query(Person).filter(Person.building == g.current_building).all()
for p in persons:
if p.groups.count():
print("%s have groups" % p)
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