USE AdventureWorks2012;
GO
SELECT
f.name AS foreign_key_name
,OBJECT_NAME(f.parent_object_id) AS table_name
,COL_NAME(fc.parent_object_id, fc.parent_column_id) AS constraint_column_name
,OBJECT_NAME (f.referenced_object_id) AS referenced_object
,COL_NAME(fc.referenced_object_id, fc.referenced_column_id) AS referenced_column_name
,is_disabled
,delete_referential_action_desc
,update_referential_action_desc
FROM sys.foreign_keys AS f
INNER JOIN sys.foreign_key_columns AS fc
ON f.object_id = fc.constraint_object_id
WHERE f.parent_object_id = OBJECT_ID('HumanResources.Employee');
转自:http://msdn.microsoft.com/zh-cn/library/ms190196.aspx