Tuesday, August 23, 2016

-- query to determine to find all the keys in the database, which columns are involved, and report if they are combined keys (ColumnCount)


;with CTE_SingleKey as 
(select
'Single Column Key' as KeyType
,COUNT(scu.column_name) as ColumnCount
,scu.CONSTRAINT_NAME
,kc.type
,kc.type_desc
,scu.TABLE_NAME
from sys.key_constraints kc
join INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE scu
on kc.name = scu.CONSTRAINT_NAME
group by
scu.CONSTRAINT_NAME
,kc.type
,kc.type_desc
,scu.TABLE_NAME
having COUNT(scu.column_name)
)
,CTE_MultiKey as 
(select
'Multi Column Key' as KeyType
,COUNT(scu.column_name) as ColumnsCount
,scu.CONSTRAINT_NAME
,kc.type
,kc.type_desc
,scu.TABLE_NAME
from sys.key_constraints kc
join INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE scu
on kc.name = scu.CONSTRAINT_NAME
group by
scu.CONSTRAINT_NAME
,kc.type
,kc.type_desc
,scu.TABLE_NAME
having COUNT(scu.column_name) > 1
)

select mkey.*
,scu.COLUMN_NAME
from CTE_MultiKey mkey
join INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE scu
on mkey.CONSTRAINT_NAME = scu.CONSTRAINT_NAME

union

select skey.*
,scu.COLUMN_NAME
from CTE_SingleKey skey
join INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE scu
on skey.CONSTRAINT_NAME = scu.CONSTRAINT_NAME
order by
TABLE_NAME
,CONSTRAINT_NAME
,COLUMN_NAME



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