Question

I have a Django-created tables in PostgreSQL 8.4 database, where one table “extends” another. One table (FooPayment) has primary key, which references another table (Payment). In SQL it looks like this:

CREATE TABLE foo.payments_payment
(
  id integer NOT NULL DEFAULT nextval('payments_payment_id_seq'::regclass),
  user_id integer NOT NULL,
  ...
  CONSTRAINT payments_payment_pkey PRIMARY KEY (id),
  CONSTRAINT payments_payment_user_id_fkey FOREIGN KEY (user_id)
      REFERENCES auth.auth_user (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION DEFERRABLE INITIALLY DEFERRED
)

CREATE TABLE foo.payments_foopayment
(
  payment_ptr_id integer NOT NULL,
  ...
  CONSTRAINT payments_foopayment_pkey PRIMARY KEY (payment_ptr_id),
  CONSTRAINT payments_foopayment_payment_ptr_id_fkey FOREIGN KEY (payment_ptr_id)
      REFERENCES foo.payments_payment (id) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION DEFERRABLE INITIALLY DEFERRED,
  ...
)

However, I’ve not to use Django ORM for various reasons, and I’m trying to access the tables from SQLAlchemy (I’m using version 0.6.6, as was installed with pip):

# Base = declarative_base()
...

class Payment(Base):
    __tablename__ = 'payments_payment'
    __table_args__ = {'schema': 'foo', 'autoload': True}
    user = relation(User, backref='payments')

class FooPayment(Payment):
    __tablename__ = 'payments_foopayment'
    __table_args__ = {'schema': 'foo', 'autoload': True}

When I’m doing this as superuser, everything works. When I’m connecting as low-privileged user I get an exception:

Traceback (most recent call last):
  File "./test.py", line 3, in <module>
    from foos import models
  File "./foos/models.py", line 127, in <module>
    class FooPayment(Payment):
  File "lib/python2.6/site-packages/sqlalchemy/ext/declarative.py", line 1167, in __init__
    _as_declarative(cls, classname, cls.__dict__)
  File "lib/python2.6/site-packages/sqlalchemy/ext/declarative.py", line 1099, in _as_declarative
    ignore_nonexistent_tables=True)
  File "lib/python2.6/site-packages/sqlalchemy/sql/util.py", line 260, in join_condition
    "between '%s' and '%s'.%s" % (a.description, b.description, hint))
sqlalchemy.exc.ArgumentError: Can't find any foreign key relationships between 'payments_payment' and 'payments_foopayment'.

When I’m connecting as this low-privileged user with PgAdmin3, I see the relationship in GUI. I can also see it with this statement, SQLAlchemy issues itself:

SELECT conname, pg_catalog.pg_get_constraintdef(oid, true) as condef
    FROM  pg_catalog.pg_constraint r
    WHERE r.conrelid = 16234 AND r.contype = 'f'
    ORDER BY 1

Which properly returns a row, containing

"payments_foopayment_payment_ptr_id_fkey"; "FOREIGN KEY (payment_ptr_id) REFERENCES payments_payment(id) DEFERRABLE INITIALLY DEFERRED"

As for database permissions, both payments_payment and payments_foopayment are GRANTed SELECT and UPDATE. I’ve tried temporarily granting all permissions on them, without any success. If this matters, a seqence payments_payment_id_seq is GRANTed for SELECT and USAGE. Obviously, schema foo is GRANTed for USAGE.

How should I either define the relationship manually in Python, or do something on DB side, so introspection would work for non-privileged user?

Hints on debugging the problem are also very much welcomed, as I’m completely lost in SA internals.

Answer 1

You can log the queries of SQLAlchemy, and compare what happens with different users.

import logging

# Early in your main()
logging.basicConfig()
logging.getLogger('sqlalchemy.engine').setLevel(logging.INFO)
logging.getLogger('sqlalchemy.orm').setLevel(logging.INFO)

logging.DEBUG logs the response data as well.

As far as I can tell, for reflection, SQLAlchemy uses table OIDs and queries pg_catalog; you gave an example. The code is in SQLAlchemy.dialects.postgresql.base.

If the permissions for autoload give you grief, you can declare the relationship in code with something like this:

class FooPayment(Payment):
    payment_ptr_id = Column(Integer, ForeignKey(Payment.id), primary_key=True)
    payment = relationship(
                  Payment, foreign_keys=[payment_ptr_id], backref='foo_payment')