Reputation: 8068
I'm having problems with SQLAlchemy's select_from
statement when using the core component. I try to construct an outer join query which currently looks like:
query = select([b1.c.id, b1.c.num, n1.c.name, n1.c.num, ...]
).where(and_(
... some conditions ...
)
).select_from(
???.outerjoin(
n1,
and_(
... some conditions ...
)
).select_from(... more outer joins similar to the above ...)
According to the docs, the structure should look like this:
table1 = table('t1', column('a'))
table2 = table('t2', column('b'))
s = select([table1.c.a]).\
select_from(
table1.join(table2, table1.c.a==table2.c.b)
)
My problem is that I don't have a table1 object in this case, as the select ...
part consists of columns and not a single table (see question marks in my query). I've tried using n1.outerjoin(n1...
, but that caused an exception (Exception: (ProgrammingError) table name "n1" specified more than once
).
The above snippet is derived from a working session-based (ORM) query, which I try to convert (with limited success).
b = Table('b', metadata,
Column('id', Integer, Sequence('seq_b_id')),
Column('num', Integer, nullable=False),
Column('active', Boolean, default=False),
)
n = Table('n', metadata,
Column('b_id', Integer, nullable=False),
Column('num', Integer, nullable=False),
Column('active', Boolean, default=False),
)
p = Table('p', metadata,
Column('b_id', Integer, nullable=False),
Column('num', Integer, nullable=False),
Column('active', Boolean, default=False),
)
n1 = aliased(n, name='n1')
n2 = aliased(n, name='n2')
b1 = aliased(b, name='b1')
b2 = aliased(b, name='b2')
p1 = aliased(p, name='p1')
p2 = aliased(p, name='p2')
result = sess.query(b1.id, b1.num, n1.c.name, n1.c.num, p1.par, p1.num).filter(
b1.active==False,
b1.num==sess.query(func.max(b2.num)).filter(
b2.id==b1.id
)
).outerjoin(
n1,
and_(
n1.c.b_id==b1.id,
n1.c.num<=num,
n1.c.active==False,
n1.c.num==sess.query(func.max(n2.num)).filter(
n2.id==n1.c.id
)
)
).outerjoin(
p1,
and_(
p1.b_id==b1.id,
p1.num<=num,
p1.active==False,
p1.num==sess.query(func.max(p2.num)).filter(
p2.id==p1.id
)
)
).order_by(b1.id)
How do I go about converting this ORM query into a plain Core query? Update:
I was able to narrow down the problem. It seems that a combination of two select_from
calls causes the problem.
customer = Table('customer', metadata,
Column('id', Integer),
Column('name', String(50)),
)
order = Table('order', metadata,
Column('id', Integer),
Column('customer_id', Integer),
Column('order_num', Integer),
)
address = Table('address', metadata,
Column('id', Integer),
Column('customer_id', Integer),
Column('city', String(50)),
)
metadata.create_all(db)
customer1 = aliased(customer, name='customer1')
order1 = aliased(order, name='order1')
address1 = aliased(address, name='address1')
columns = [
customer1.c.id, customer.c.name,
order1.c.id, order1.c.order_num,
address1.c.id, address1.c.city
]
query = select(columns)
query = query.select_from(
customer1.outerjoin(
order1,
and_(
order1.c.customer_id==customer1.c.id,
)
)
)
query = query.select_from(
customer1.outerjoin(
address1,
and_(
customer1.c.id==address1.c.customer_id
)
)
)
result = connection.execute(query)
for r in result.fetchall():
print r
The above code causes the following exception:
ProgrammingError: (ProgrammingError) table name "customer1" specified more than once
'SELECT customer1.id, customer.name, order1.id, order1.order_num, address1.id, address1.city \nFROM customer, customer AS customer1 LEFT OUTER JOIN "order" AS order1 ON order1.customer_id = customer1.id, customer AS customer1 LEFT OUTER JOIN address AS address1 ON customer1.id = address1.customer_id' {}
If I was a bit more experienced in using SQLAlchemy, I would say this could be a bug...
Upvotes: 4
Views: 1926
Reputation: 8068
I finally managed to solved the problem. Instead of cascading select_from
, additional joins need to be chained to the actual join. The above query would read:
query = select(columns)
query = query.select_from(
customer1.outerjoin(
order1,
and_(
order1.c.customer_id==customer1.c.id,
)
).outerjoin(
address1,
and_(
customer1.c.id==address1.c.customer_id
)
)
)
Upvotes: 3