How to Find the Foreign Key Associated with a Given Primary Key

Question

All, I want a way to establish which columns in a given database are joined via PK/FK relationships. I can return the PK/FK information for a given table via

SELECT *  
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS cu 
WHERE EXISTS (
    SELECT tc.* 
    FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS AS tc 
    WHERE tc.CONSTRAINT_CATALOG = 'MyDatabase'  
        AND tc.TABLE_NAME = 'MyTable'  
        /*AND tc.CONSTRAINT_TYPE = 'PRIMARY KEY'*/
        AND tc.CONSTRAINT_NAME = cu.CONSTRAINT_NAME);
GO

but for a PK returned from such a query how do I establish the ascociated FK (assuming there is one)?

Edit. I know you can also get the referenced tables via

SELECT CONSTRAINT_NAME = name, 
       FOREIGN_SCHEMA = OBJECT_SCHEMA_NAME(parent_object_id), 
       FOREIGN_TABLE = OBJECT_NAME(parent_object_id), 
       REFERENCED_SCHEMA = OBJECT_SCHEMA_NAME(referenced_object_id), 
       REFERENCED_TABLE = OBJECT_NAME(referenced_object_id) 
FROM sys.foreign_keys
WHERE OBJECT_NAME(referenced_object_id) = 'MyTable';
GO

but I am struggling now to get the explicit column references.

Thanks for your time.

Answer 1

Here's a simple query to match up foreign keys to their referenced tables/columns:

SELECT
    o1.name AS FK_table,
    c1.name AS FK_column,
    fk.name AS FK_name,
    o2.name AS PK_table,
    c2.name AS PK_column,
    pk.name AS PK_name,
    fk.delete_referential_action_desc AS Delete_Action,
    fk.update_referential_action_desc AS Update_Action
FROM sys.objects o1
    INNER JOIN sys.foreign_keys fk
        ON o1.object_id = fk.parent_object_id
    INNER JOIN sys.foreign_key_columns fkc
        ON fk.object_id = fkc.constraint_object_id
    INNER JOIN sys.columns c1
        ON fkc.parent_object_id = c1.object_id
        AND fkc.parent_column_id = c1.column_id
    INNER JOIN sys.columns c2
        ON fkc.referenced_object_id = c2.object_id
        AND fkc.referenced_column_id = c2.column_id
    INNER JOIN sys.objects o2
        ON fk.referenced_object_id = o2.object_id
    INNER JOIN sys.key_constraints pk
        ON fk.referenced_object_id = pk.parent_object_id
        AND fk.key_index_id = pk.unique_index_id
ORDER BY o1.name, o2.name, fkc.constraint_column_id

The output has eight columns: the table and column names for the foreign keys (FK_table, FK_column), the names of the foreign-key constraints (FK_name), the referenced PK or unique index table and column names (PK_table, PK_column), the name of the referenced PK or unique index (PK_name), and the update/delete cascade actions (Delete_Action, Update_Action).

(Edited to add some more output columns.)

Answer 2

This query nets you all of the FK relationships in the database - FK constraint name, schema/table of referencing table, referencing column name, schema/table of referenced table, and referenced column name. There will be multiple rows for a multi-column constraint.

SELECT 
    FK = OBJECT_NAME(pt.constraint_object_id),
    Referencing_table = QUOTENAME(OBJECT_SCHEMA_NAME(pt.parent_object_id))
            + '.' + QUOTENAME(OBJECT_NAME(pt.parent_object_id)),
    Referencing_col = QUOTENAME(pc.name), 
    Referenced_table = QUOTENAME(OBJECT_SCHEMA_NAME(pt.referenced_object_id)) 
            + '.' + QUOTENAME(OBJECT_NAME(pt.referenced_object_id)),
    Referenced_col = QUOTENAME(rc.name)
FROM sys.foreign_key_columns AS pt
INNER JOIN sys.columns AS pc
ON pt.parent_object_id = pc.[object_id]
AND pt.parent_column_id = pc.column_id
INNER JOIN sys.columns AS rc
ON pt.referenced_column_id = rc.column_id
AND pt.referenced_object_id = rc.[object_id]
ORDER BY Referencing_table, FK, pt.constraint_column_id;

If you are after the columns from a specific primary key constraint, and you already know the name of that PK constraint, you can write this:

DECLARE @PK_Constraint SYSNAME = N'Name of PK constraint';

SELECT
    FK = OBJECT_NAME(fkc.constraint_object_id),
    Referencing_table = QUOTENAME(OBJECT_SCHEMA_NAME(fkc.parent_object_id))
            + '.' + QUOTENAME(OBJECT_NAME(fkc.parent_object_id)),
    Referencing_col = QUOTENAME(pc.name), 
    Referenced_table = QUOTENAME(OBJECT_SCHEMA_NAME(fkc.referenced_object_id)) 
            + '.' + QUOTENAME(OBJECT_NAME(fkc.referenced_object_id)),
    Referenced_col = QUOTENAME(rc.name)
FROM sys.foreign_key_columns AS fkc
INNER JOIN sys.columns AS pc
ON fkc.parent_object_id = pc.[object_id]
AND fkc.parent_column_id = pc.column_id
INNER JOIN sys.columns AS rc
ON fkc.referenced_column_id = rc.column_id
AND fkc.referenced_object_id = rc.[object_id]
WHERE EXISTS 
(
  SELECT 1 FROM sys.indexes AS i
  INNER JOIN sys.foreign_keys AS fk
  ON i.[object_id] = fk.referenced_object_id
  AND i.index_id = fk.key_index_id
  AND fk.[object_id] = fkc.constraint_object_id
  AND i.name = @PK_Constraint
)
ORDER BY Referencing_table, FK, fkc.constraint_column_id;

If you just want to include the PK name along with the other information:

SELECT 
    FK = OBJECT_NAME(fkc.constraint_object_id),
    Referencing_table = QUOTENAME(OBJECT_SCHEMA_NAME(fkc.parent_object_id))
            + '.' + QUOTENAME(OBJECT_NAME(fkc.parent_object_id)),
    Referencing_col = QUOTENAME(pc.name),
    Referenced_table = QUOTENAME(OBJECT_SCHEMA_NAME(fkc.referenced_object_id)) 
            + '.' + QUOTENAME(OBJECT_NAME(fkc.referenced_object_id)),
    Referenced_col = QUOTENAME(rc.name),
    PK = pk.name
FROM sys.foreign_key_columns AS fkc
INNER JOIN sys.columns AS pc
ON fkc.parent_object_id = pc.[object_id]
AND fkc.parent_column_id = pc.column_id
INNER JOIN sys.columns AS rc
ON fkc.referenced_column_id = rc.column_id
AND fkc.referenced_object_id = rc.[object_id]
INNER JOIN (SELECT i.name, fk.[object_id]
  FROM sys.indexes AS i
  INNER JOIN sys.foreign_keys AS fk
  ON i.[object_id] = fk.referenced_object_id
  AND i.index_id = fk.key_index_id
) AS pk
ON pk.[object_id] = fkc.constraint_object_id
ORDER BY Referencing_table, FK, fkc.constraint_column_id;

There are also tricks to getting the column list in, say, a comma-separated list or individual columns, instead of being spread across rows, but I'm not going to invest in modifying these queries to produce that until I know exactly which form you're after.