Programmatically picking apart foreign-key constraints

Another quickie today. Don't ask why, but this morning I was trying to programmatically work out which tables have foreign-key references to table Sales, because they're preventing fast partition switching on table Sales.

After more spelunking around I thought I had it with this code:

SELECT * FROM sys.all_objects
WHERE [type] = 'F' AND [parent_object_id] = OBJECT_ID ('Sales');
GO

But that gives me the foreign-key references FROM the Sales table, not TO the Sales table.

Eventually I found the sys.foreign_key_columns table which does the trick, using the following code:

SELECT OBJECT_NAME ([parent_object_id]) AS 'Referencing Table', * FROM sys.foreign_key_columns
WHERE [referenced_object_id] = OBJECT_ID ('Sales');
GO

And it also gives me the column IDs in both tables involved in the constraint. 

Enjoy!

6 thoughts on “Programmatically picking apart foreign-key constraints

  1. Very cool. You coerced me into spelunking furhter as to how how I might quickly peruse all relationships for annomolies. This did it for me.

    /*
    Robert Magrogan, Robert@MyRamLink.com October 9 2010
    Ref: SQLSkills and Michael Stwart http://www.sqlskills.com/blogs/paul/post/Programmatically-picking-apart-foreign-key-constraints.aspx
    Get all relationships (Foreign keys contraingts) for a given table
    */
    SELECT
    OBJECT_NAME(fkc.referenced_object_id) AS ReferencedObject
    , COL_NAME(fkc.referenced_object_id , fkc.referenced_column_id) AS ‘ReferencedColumn’
    , COL_NAME(fkc.parent_object_id , fkc.parent_column_id) AS ‘ReferencingColumnName’
    , OBJECT_NAME(fkc.parent_object_id) AS ‘ReferencingObjectName’
    , OBJECT_NAME(fkc.constraint_object_id) AS ‘ForeignKeyName’
    , FK.is_not_for_replication
    , FK.delete_referential_action_desc
    , fk.update_referential_action_desc
    , fk.schema_id
    FROM sys.foreign_key_columns FKC
    INNER JOIN sys.foreign_keys FK ON FKC.constraint_object_id=FK.object_id
    INNER JOIN sys.schemas S ON FK.schema_id=S.schema_id
    –WHERE [referenced_object_id]=OBJECT_ID(‘tblPersons’)
    ORDER BY 1,2,4,5

    –SELECT TOP 2 * FROM sys.schemas
    –SELECT TOP 2 * FROM sys.foreign_key_columns
    –SELECT TOP 2 * FROM sys.foreign_keys

  2. Here’s my variation on Robert’s script to include the schema name in both referenced and referencing columns:

    SELECT rs.name + ‘.’ + r.name AS ‘This table’,
    rc.name AS ‘and Column’,
    ps.name + ‘.’ + p.name AS ‘Is referenced by this table’,
    pc.name AS ‘and by this Column’,
    OBJECT_NAME(fk.constraint_object_id) AS ‘in Constraint’
    FROM sys.foreign_key_columns fk

    join sys.objects p
    on fk.parent_object_id = p.object_id
    join sys.schemas ps
    on p.schema_id = ps.schema_id
    join sys.columns pc
    on fk.parent_object_id = pc.object_id
    and fk.parent_column_id = pc.column_id

    join sys.objects r
    on fk.referenced_object_id = r.object_id
    join sys.schemas rs
    on r.schema_id = rs.schema_id
    join sys.columns rc
    on fk.referenced_object_id = rc.object_id
    and fk.referenced_column_id = rc.column_id

    WHERE fk.referenced_object_id = OBJECT_ID (‘wf.department’)
    –WHERE rs.name = ‘WF’ — to see all foreign key relationships for a schema

    order by rs.name, r.name, rc.name, ps.name, p.name;
    GO

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