Reputation: 3
SELECT
maintener.*,
(SELECT COUNT(*)
FROM device d
WHERE d.in_stock_maintener_id = maintener.id) AS in_stock_devices
FROM maintener;
I'm creating a report that show all mainteners but i need to show the number of devices that each one of that mainteners has by looking at the devices model reference in_stock_maintener_id;
I have this models in my persist sqlalchemy.
class Maintener(persist.Base):
__tablename__ = 'maintener'
id = Column(Integer, primary_key=True)
name = Column(String(255))
document_number = Column(String(30))
phone_1 = Column(String(12))
phone_2 = Column(String(12))
email = Column(String(255))
class Device(persist.Base):
__tablename__ = 'device'
id = Column(Integer, primary_key=True)
serial = Column(String(45))
in_stock = Column(SmallInteger)
in_stock_maintener_id = Column(ForeignKey(u'maintener.id'), nullable=True, index=True)
in_stock_maintener = relationship(u'Maintener', lazy='noload', \
primaryjoin='Device.in_stock_maintener_id == Maintener.id')
If anyone could help me, i'll be grateful =)
Upvotes: 0
Views: 104
Reputation: 76952
sq = (
session
.query(func.count())
.select_from(Device)
.filter(Device.in_stock_maintener_id == Maintener.id)
).as_scalar()
q = session.query(Maintener, sq.label('in_stock_devices'))
Query above will return an enumerable of tuple(Maintener, Integer)
.
If you would like to have columns instead (as per your comment), then you can either specify the columns you want in the query implicitly:
q = session.query(Maintener.id, Maintener.name, sq.label('in_stock_devices'))
or if you would like all columns (as in SELECT *
), then you could query the Table
instead of the mapped entity:
q = session.query(Maintener.__table__, sq.label('in_stock_devices'))
Above I assumed that you use declarative
extension.
Upvotes: 1