-- SPDX-License-Identifier: Apache-2.0

/*
 * Copyright (c) 2019-2025. Ivan Vakhrushev and others.
 * https://github.com/mfvanek/pg-index-health-sql
 *
 * Licensed under the Apache License 2.0
 */

-- Finds completely identical foreign keys
--
-- Based on a query from https://habr.com/ru/articles/803841/
with
    fk_with_attributes as (
        select
            c.conrelid as table_oid,
            c.confrelid as foreign_table_oid,
            u.attposition,
            col.attnotnull,
            quote_ident(col.attname) as column_name,
            quote_ident(c.conname) as constraint_name
        from
            pg_catalog.pg_constraint c
            inner join pg_catalog.pg_namespace nsp on nsp.oid = c.connamespace
            inner join lateral unnest(c.conkey) with ordinality u(attnum, attposition) on true
            inner join pg_catalog.pg_attribute col on col.attrelid = c.conrelid and col.attnum = u.attnum
        where
            c.contype = 'f' and
            c.conparentid = 0 and c.coninhcount = 0 and /* not a constraint in a partition */
            nsp.nspname = 'public' /* TODO */
    ),

    fk_with_attributes_grouped as (
        select
            constraint_name,
            table_oid,
            foreign_table_oid,
            array_agg(column_name || ',' || attnotnull::text order by attposition) as columns
        from fk_with_attributes
        group by constraint_name, table_oid, foreign_table_oid
    )

select
    c1.table_oid::regclass::text as table_name,
    c1.foreign_table_oid::regclass::text as target_table_name,
    ARRAY[c1.constraint_name, c2.constraint_name]
from
    fk_with_attributes_grouped c1
    inner join fk_with_attributes_grouped c2
        on c2.constraint_name > c1.constraint_name and /* to prevent duplicated rows in output */
        c2.table_oid = c1.table_oid and
        c2.foreign_table_oid = c1.foreign_table_oid and
        c2.columns = c1.columns
order by table_name, c1.constraint_name, c2.constraint_name;
