在 SQL Server 中，如何从外键中获取引用的表 + 列名？
注意：不是键所在的表/列，而是它所引用的键。
例子：
当 key[FA_MDT_ID]在 table 中时[T_ALV_Ref_FilterDisplay]。指[T_AP_Ref_Customer].[MDT_ID]
例如在创建这样的约束时：
ALTER TABLE [dbo].[T_ALV_Ref_FilterDisplay] WITH CHECK ADD CONSTRAINT [FK_T_ALV_Ref_FilterDisplay_T_AP_Ref_Customer] FOREIGN KEY([FA_MDT_ID])
REFERENCES [dbo].[T_AP_Ref_Customer] ([MDT_ID])
GO
作为输入，我需要[T_AP_Ref_Customer].[MDT_ID]
得到[T_ALV_Ref_FilterAnzeige].[FA_MDT_ID]
没关系，这是正确的答案： 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
注意：
Information_schema 不包含索引（它确实找到了唯一约束）。
因此，如果您想根据唯一索引查找外键，则需要查看 microsoft 专有表：
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
边缘情况的证明测试：
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
如果您可以使用特定于 SQL Server 的架构目录视图，则此查询将返回您要查找的内容：
SELECT
fk.name,
OBJECT_NAME(fk.parent_object_id) 'Parent table',
c1.name 'Parent column',
OBJECT_NAME(fk.referenced_object_id) 'Referenced table',
c2.name 'Referenced column'
FROM
sys.foreign_keys fk
INNER JOIN
sys.foreign_key_columns fkc ON fkc.constraint_object_id = fk.object_id
INNER JOIN
sys.columns c1 ON fkc.parent_column_id = c1.column_id AND fkc.parent_object_id = c1.object_id
INNER JOIN
sys.columns c2 ON fkc.referenced_column_id = c2.column_id AND fkc.referenced_object_id = c2.object_id
不确定如何 - 如果有的话 - 您可以从 INFORMATION_SCHEMA 视图中获得相同的信息....
我想要一个版本，它可以让我找到所有具有/缺少约束的“Key”和“ID”列。所以我想要所有列与所有 PK OR FK OR Null 的列表相比，这是我的查询。希望它可以帮助别人！
SELECT
c.table_schema
,c.table_name
,c.column_name
,KeyConstraints.constraint_type
,KeyConstraints.constraint_schema
,KeyConstraints.constraint_name
,KeyConstraints.referenced_table_schema
,KeyConstraints.referenced_table_name
,KeyConstraints.referenced_column_name
,KeyConstraints.update_rule
,KeyConstraints.delete_rule
FROM information_schema.columns AS c
LEFT JOIN
(
SELECT
FK.table_schema AS TABLE_SCHEMA
,FK.table_name
,CU.column_name
,FK.constraint_type
,c.constraint_schema
,C.constraint_name
,PK.table_schema AS REFERENCED_TABLE_SCHEMA
,PK.table_name AS REFERENCED_TABLE_NAME
,CCU.column_name AS REFERENCED_COLUMN_NAME
,C.update_rule
,C.delete_rule
FROM information_schema.referential_constraints AS C
INNER JOIN information_schema.table_constraints AS FK
ON C.constraint_name = FK.constraint_name
INNER JOIN information_schema.table_constraints AS PK
ON C.unique_constraint_name = PK.constraint_name
INNER JOIN information_schema.key_column_usage AS CU
ON C.constraint_name = CU.constraint_name
INNER JOIN information_schema.constraint_column_usage AS CCU
ON PK.constraint_name = CCU.constraint_name
WHERE ( FK.constraint_type = 'FOREIGN KEY' )
UNION
SELECT
ccu.table_schema
,ccu.table_name
,ccu.column_name
,tc.constraint_type
,ccu.constraint_schema
,ccu.constraint_name
,NULL
,NULL
,NULL
,NULL
,NULL
FROM information_schema.constraint_column_usage ccu
INNER JOIN information_schema.table_constraints tc
ON ccu.table_schema = tc.table_schema
AND ccu.table_name = tc.table_name
WHERE tc.constraint_type = 'PRIMARY KEY'
) AS KeyConstraints
ON c.table_schema = KeyConstraints.table_schema
AND c.table_name = KeyConstraints.table_name
AND c.column_name = KeyConstraints.column_name
WHERE c.column_name LIKE '%ID' OR c.column_name LIKE '%Key'
ORDER BY c.table_schema
,c.table_name
,c.column_name
;
you can use the following script in order to find all the fk,pk relationship for specific table
*DECLARE @tablename VARCHAR(100)
SET @tablename='xxxxxxx'
Select 'Referenced by FK table' AS Type, FK.TABLE_SCHEMA, FK.TABLE_NAME AS
'FK_TABLE_NAME' ,cu.COLUMN_NAME AS 'FK_ReferencingColumn',PK.TABLE_NAME AS
'PK_TABLE_NAME',
ku.COLUMN_NAME AS 'PK_ReferencedColumn'
From INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS As RC
Join INFORMATION_SCHEMA.TABLE_CONSTRAINTS As PK
On PK.CONSTRAINT_NAME = RC.UNIQUE_CONSTRAINT_NAME
Join INFORMATION_SCHEMA.TABLE_CONSTRAINTS As FK
On FK.CONSTRAINT_NAME = RC.CONSTRAINT_NAME
JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE cu
ON cu.CONSTRAINT_NAME = Rc.CONSTRAINT_NAME
JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE ku
ON ku.CONSTRAINT_NAME = RC.UNIQUE_CONSTRAINT_NAME
Where
PK.TABLE_NAME = @tablename
UNION
SELECT 'Referencing PK table' AS Type, FK.TABLE_SCHEMA, FK.TABLE_NAME AS
'FK_TABLE_NAME' ,cu.COLUMN_NAME AS 'FK_ReferencingColumn',PK.TABLE_NAME AS
'PK_TABLE_NAME',
ku.COLUMN_NAME AS 'PK_ReferencedColumn'
From INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS As RC
Join INFORMATION_SCHEMA.TABLE_CONSTRAINTS As PK
On PK.CONSTRAINT_NAME = RC.UNIQUE_CONSTRAINT_NAME
Join INFORMATION_SCHEMA.TABLE_CONSTRAINTS As FK
On FK.CONSTRAINT_NAME = RC.CONSTRAINT_NAME
JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE cu
ON cu.CONSTRAINT_NAME = Rc.CONSTRAINT_NAME
JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE ku
ON ku.CONSTRAINT_NAME = RC.UNIQUE_CONSTRAINT_NAME
Where
fk.TABLE_NAME = @tablename*