Sqlserver 2005+：查看索引【index】的【使用情况】：无效的索引、高成本索引

set nocount on
go

--无效索引排序
select top 20
db_name() as db_name
,schema_name(o.schema_id) as schema_name
,o.name as object_name
,i.name as index_name,i.is_unique_constraint,(case i.index_id when 1 then 'is_clustered' else '' end) as is_clustered
,ir.rowcnt
,s.user_updates
,(s.user_lookups + s.user_scans + s.user_seeks) as [retrieval usage] --查询使用次数
,s.system_seeks + s.system_scans + s.system_lookups as [system usage]
,'DROP INDEX ' + quotename(o.name) + '.' + quotename(i.name) as [ -- dsql]
from sys.dm_db_index_usage_stats s
inner join sys.indexes i on s.database_id=db_id() and s.[object_id] = i.[object_id] and s.index_id = i.index_id
inner join sys.objects o on i.object_id = o.object_id
inner join sysindexes ir on ir.id=i.object_id and ir.indid=i.index_id
where s.database_id = db_id()
and o.is_ms_shipped = 0
and i.name is not null
--and s.user_seeks = 0
--and s.user_scans = 0
--and s.user_lookups = 0
and (s.user_lookups + s.user_scans + s.user_seeks)=0
order by s.user_updates desc

go

-- 维护成本 排序
select top 20
db_name() as db_name
,schema_name(o.schema_id) as schema_name
,o.name as object_name
,i.name as index_name,i.is_unique_constraint,(case i.index_id when 1 then 'is_clustered' else '' end) as is_clustered
,ir.rowcnt
,s.user_updates -- as [update usage] --更新成本
,(s.user_seeks + s.user_scans + s.user_lookups) as [retrieval usage] --查询使用次数
,(s.user_updates) - (s.user_seeks + user_scans + s.user_lookups) as [maintenance cost] --用户维护成本
,s.system_seeks + s.system_scans + s.system_lookups as [system usage] --系统内部维护次数，--内部维护成本
,s.last_user_seek
,s.last_user_scan
,s.last_user_lookup
,'DROP INDEX ' + quotename(o.name) + '.' + quotename(i.name) as [ -- dsql]
from sys.dm_db_index_usage_stats s
inner join sys.indexes i on s.database_id=db_id() and s.[object_id] = i.[object_id] and s.index_id = i.index_id
inner join sys.objects o on i.object_id = o.object_id
inner join sysindexes ir on ir.id=i.object_id and ir.indid=i.index_id
where s.database_id = db_id()
and o.is_ms_shipped = 0
and i.name is not null
and (s.user_seeks + s.user_scans + s.user_lookups) > 0
order by [maintenance cost] desc