Generate sql with subquery as a column in select statement using SQLAlchemy

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Question :

Generate sql with subquery as a column in select statement using SQLAlchemy

Is there a way to make SQLAlchemy generate a query with a custom column that is a subquery that correlates with current row:

     SELECT count(1) FROM tab2 
     WHERE tab2.tab1_id =
     GROUP BY tab2.col1
 ) as cnt
FROM tab1

using the ORM API?

session.query(Tab1, ?(subquery for additional column)?).filter(...).limit(100)

I’m using PostgreSQL 9.3 and old version of SQLAlchemy 0.9.8

Asked By: Antigluk


Answer #1:

If you need this often, and/or the count is an integral part of your Tab1 model, you should use a hybrid property such as described in the other answer. If on the other hand you need this just for a single query, then you could just create the scalar subquery using Query.label(), or Query.as_scalar():

count_stmt = session.query(func.count(1)).
    filter(Tab2.tab1_id ==

session.query(Tab1, count_stmt).filter(...).limit(100)

The subquery will automatically correlate what it can from the enclosing query.

Answered By: Antigluk

Answer #2:

You can do this, but it works in a quite different way to how you have written it. You can create a property of Tab1 which depends on the relationship to tab2 (assuming that tab2.tab1_id is a foreign key, which it should be.

Your models look like this:

class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    children = relationship("Child")

class Child(Base):
    __tablename__ = 'child'
    id = Column(Integer, primary_key=True)
    parent_id = Column(Integer, ForeignKey(''))

as per the docs on relationships

Then you can add something like

def number_of_children(self):
    if self.children:
        return len(self.children)
    return 0

def number_of_children(cls):
    return (select([func.count(])
            .where(Child.cover_id ==

to the Parent model, as per this answer and more docs.

Once you’ve done this, you can filter on this property the same as any other column-based one.

Answered By: Ilja Everilä

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