Peu importe, c'est la bonne réponse :
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
Remarque :
Information_schema ne contient pas d'index (il trouve des contraintes uniques).
Donc, si vous souhaitez trouver des clés étrangères basées sur des indices uniques, vous devez parcourir les tables propriétaires de Microsoft :
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
Test de preuve pour les cas extrêmes :
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
