कोई बात नहीं, यह सही उत्तर है:
http://msdn.microsoft.com/en-us/library/aa175805(SQL.80).aspx
SELECT
KCU1.CONSTRAINT_SCHEMA AS FK_CONSTRAINT_SCHEMA
,KCU1.CONSTRAINT_NAME AS FK_CONSTRAINT_NAME
,KCU1.TABLE_SCHEMA AS FK_TABLE_SCHEMA
,KCU1.TABLE_NAME AS FK_TABLE_NAME
,KCU1.COLUMN_NAME AS FK_COLUMN_NAME
,KCU1.ORDINAL_POSITION AS FK_ORDINAL_POSITION
,KCU2.CONSTRAINT_SCHEMA AS REFERENCED_CONSTRAINT_SCHEMA
,KCU2.CONSTRAINT_NAME AS REFERENCED_CONSTRAINT_NAME
,KCU2.TABLE_SCHEMA AS REFERENCED_TABLE_SCHEMA
,KCU2.TABLE_NAME AS REFERENCED_TABLE_NAME
,KCU2.COLUMN_NAME AS REFERENCED_COLUMN_NAME
,KCU2.ORDINAL_POSITION AS REFERENCED_ORDINAL_POSITION
FROM INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS AS RC
INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS KCU1
ON KCU1.CONSTRAINT_CATALOG = RC.CONSTRAINT_CATALOG
AND KCU1.CONSTRAINT_SCHEMA = RC.CONSTRAINT_SCHEMA
AND KCU1.CONSTRAINT_NAME = RC.CONSTRAINT_NAME
INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS KCU2
ON KCU2.CONSTRAINT_CATALOG = RC.UNIQUE_CONSTRAINT_CATALOG
AND KCU2.CONSTRAINT_SCHEMA = RC.UNIQUE_CONSTRAINT_SCHEMA
AND KCU2.CONSTRAINT_NAME = RC.UNIQUE_CONSTRAINT_NAME
AND KCU2.ORDINAL_POSITION = KCU1.ORDINAL_POSITION
नोट:
Information_schema में इंडेक्स नहीं हैं (इसमें अद्वितीय-अवरोध पाए जाते हैं)।
इसलिए यदि आप अद्वितीय-सूचकांकों के आधार पर विदेशी-कुंजी खोजना चाहते हैं, तो आपको Microsoft स्वामित्व तालिका पर जाना होगा:
SELECT
fksch.name AS FK_CONSTRAINT_SCHEMA
,fk.name AS FK_CONSTRAINT_NAME
,sch1.name AS FK_TABLE_SCHEMA
,t1.name AS FK_TABLE_NAME
,c1.name AS FK_COLUMN_NAME
-- The column_id is not the ordinal, it can be dropped and then there's a gap...
,COLUMNPROPERTY(c1.object_id, c1.name, 'ordinal') AS FK_ORDINAL_POSITION
,COALESCE(pksch.name,sch2.name) AS REFERENCED_CONSTRAINT_SCHEMA
,COALESCE(pk.name, sysi.name) AS REFERENCED_CONSTRAINT_NAME
,sch2.name AS REFERENCED_TABLE_SCHEMA
,t2.name AS REFERENCED_TABLE_NAME
,c2.name AS REFERENCED_COLUMN_NAME
,COLUMNPROPERTY(c2.object_id, c2.name, 'ordinal') AS REFERENCED_ORDINAL_POSITION
FROM sys.foreign_keys AS fk
LEFT JOIN sys.schemas AS fksch
ON fksch.schema_id = fk.schema_id
-- not inner join: unique indices
LEFT JOIN sys.key_constraints AS pk
ON pk.parent_object_id = fk.referenced_object_id
AND pk.unique_index_id = fk.key_index_id
LEFT JOIN sys.schemas AS pksch
ON pksch.schema_id = pk.schema_id
LEFT JOIN sys.indexes AS sysi
ON sysi.object_id = fk.referenced_object_id
AND sysi.index_id = fk.key_index_id
INNER JOIN sys.foreign_key_columns AS fkc
ON fkc.constraint_object_id = fk.object_id
INNER JOIN sys.tables AS t1
ON t1.object_id = fkc.parent_object_id
INNER JOIN sys.schemas AS sch1
ON sch1.schema_id = t1.schema_id
INNER JOIN sys.columns AS c1
ON c1.column_id = fkc.parent_column_id
AND c1.object_id = fkc.parent_object_id
INNER JOIN sys.tables AS t2
ON t2.object_id = fkc.referenced_object_id
INNER JOIN sys.schemas AS sch2
ON sch2.schema_id = t2.schema_id
INNER JOIN sys.columns AS c2
ON c2.column_id = fkc.referenced_column_id
AND c2.object_id = fkc.referenced_object_id
एज-केस के लिए प्रूफ-टेस्ट:
CREATE TABLE __groups ( grp_id int, grp_name varchar(50), grp_name2 varchar(50) )
ALTER TABLE __groups ADD CONSTRAINT UQ___groups_grp_name2 UNIQUE (grp_name2)
CREATE UNIQUE INDEX IX___groups_grp_name ON __groups(grp_name)
GO
CREATE TABLE __group_mappings( map_id int, map_grp_name varchar(50), map_grp_name2 varchar(50), map_usr_name varchar(50) )
GO
ALTER TABLE __group_mappings ADD CONSTRAINT FK___group_mappings___groups FOREIGN KEY(map_grp_name)
REFERENCES __groups (grp_name)
GO
ALTER TABLE __group_mappings ADD CONSTRAINT FK___group_mappings___groups2 FOREIGN KEY(map_grp_name2)
REFERENCES __groups (grp_name2)
GO
SELECT @@VERSION -- Microsoft SQL Server 2016 (SP1-GDR) (KB4458842)
SELECT version() -- PostgreSQL 9.6.6 on x86_64-pc-linux-gnu
GO
