Reputation: 101
So I have two tables Employee and Details like this.
class Employee(Base):
__tablename__ = 'employees'
id = Column(Integer, Sequence('employee_id_seq'), primary_key=True)
name = Column(String(50), nullable=False)
............
class Detail(Base):
__tablename__ = 'details'
id = Column(Integer, Sequence('detail_id_seq'), primary_key=True)
start_date = Column(String(50), nullable=False)
email = Column(String(50))
employee_id = Column(Integer, ForeignKey('employee.id'))
employee = relationship("Employee", backref=backref('details', order_by=id))
............
Now what I want to do is get all the employees and their corresponding details, here is what I tried.
for e, d in session.query(Employee, Detail).filter(Employee.id = Detail.employee_id).all():
print e.name, d.email
The problem with this is that it prints everything twice. I tried using .join() and also prints the results twice.
What I want to achieve is like
print Employee.name
print Employee.details.email
Upvotes: 2
Views: 2746
Reputation: 76962
If you really care only about few columns, you can specify them in the query directly:
q = session.query(Employee.name, Detail.email).filter(Employee.id == Detail.employee_id).all()
for e, d in q:
print e, d
If you do really want to load object instances, then I would do it differently:
# query all employees
q = (session.query(Employee)
# load Details in the same query
.outerjoin(Employee.details)
# let SA know that the relationship "Employee.details" is already loaded in this query so that when we access it, SA will not do another query in the database
.options(contains_eager(Employee.details))
).all()
# navigate the results simply as defined in the relationship configuration
for e in q:
print(e)
for d in e.details:
print(" ->", d)
As to your duplicate
result problem, I believe you have some "extra" in your real code which produces this error...
Upvotes: 2