app/django/db/backends/postgresql_psycopg2/introspection.py
author Sverre Rabbelier <srabbelier@gmail.com>
Fri, 12 Dec 2008 00:34:12 +0000
changeset 721 6f1d29857072
parent 323 ff1a9aa48cfd
permissions -rw-r--r--
Added a filter method to dicts The best way to explain this is probably with an example: >>> split({'foo':'bar', 'bar':['one', 'two'], 'baz': ['three', 'four']}) [{'bar': 'one', 'foo': 'bar', 'baz': 'three'}, {'bar': 'two', 'foo': 'bar', 'baz': 'three'}, {'bar': 'one', 'foo': 'bar', 'baz': 'four'}, {'bar': 'two', 'foo': 'bar', 'baz': 'four'}] Patch by: Sverre Rabbelier

from django.db.backends.postgresql.introspection import DatabaseIntrospection as PostgresDatabaseIntrospection

class DatabaseIntrospection(PostgresDatabaseIntrospection):

    def get_relations(self, cursor, table_name):
        """
        Returns a dictionary of {field_index: (field_index_other_table, other_table)}
        representing all relationships to the given table. Indexes are 0-based.
        """
        cursor.execute("""
            SELECT con.conkey, con.confkey, c2.relname
            FROM pg_constraint con, pg_class c1, pg_class c2
            WHERE c1.oid = con.conrelid
                AND c2.oid = con.confrelid
                AND c1.relname = %s
                AND con.contype = 'f'""", [table_name])
        relations = {}
        for row in cursor.fetchall():
            # row[0] and row[1] are single-item lists, so grab the single item.
            relations[row[0][0] - 1] = (row[1][0] - 1, row[2])
        return relations