SQL Server 索引重建或索引重組

查询索引的碎裂状态T-SQL语法(适用于SQL Server 2005以上):

SELECT OBJECT_NAME(dt.object_id)      ,

        si.name                        ,

        dt.avg_fragmentation_in_percent,

        dt.avg_page_space_used_in_percent

FROM

        (SELECT object_id                    ,

               index_id                    ,

               avg_fragmentation_in_percent,

               avg_page_space_used_in_percent

        FROM    sys.dm_db_index_physical_stats (DB_ID(), NULL, NULL, NULL, 'DETAILED')

        WHERE   index_id <> 0

        ) AS dt --does not return information about heaps

        INNER JOIN sys.indexes si

        ON     si.object_id = dt.object_id

           AND si.index_id  = dt.index_id



 

索引重组的时机

    *检查 Externalfragmentation 部分

         o 当avg_fragmentation_in_percent 的值介于 10 到 15 之间

    *检查 Internalfragmentation 部分

         o 当avg_page_space_used_in_percent 的值介于 60 到 75 之间

 

索引重建的时机

    *检查 Externalfragmentation 部分

         o 当avg_fragmentation_in_percent 的值大于 15

    *检查 Internalfragmentation 部分

         o 当avg_page_space_used_in_percent 的值小于 60

 

调整过的自动帮你算出哪些索引需要被重建或重组T-SQL 语法:

SELECT 'ALTER INDEX [' + ix.name + '] ON [' + s.name + '].[' + t.name + '] ' +

       CASE

              WHEN ps.avg_fragmentation_in_percent > 15

              THEN 'REBUILD'

              ELSE 'REORGANIZE'

        END +

        CASE

              WHEN pc.partition_count > 1

              THEN ' PARTITION = ' + CAST(ps.partition_number AS nvarchar(MAX))

              ELSE ''

        END,

        avg_fragmentation_in_percent

FROM    sys.indexes AS ix

        INNER JOIN sys.tables t

        ON     t.object_id = ix.object_id

        INNER JOIN sys.schemas s

        ON     t.schema_id = s.schema_id

        INNER JOIN

              (SELECT object_id                   ,

                      index_id                    ,

                       avg_fragmentation_in_percent,

                      partition_number

              FROM  sys.dm_db_index_physical_stats (DB_ID(), NULL, NULL, NULL, NULL)

              ) ps

        ON     t.object_id = ps.object_id

           AND ix.index_id = ps.index_id

        INNER JOIN

              (SELECT  object_id,

                       index_id ,

                       COUNT(DISTINCT partition_number) AS partition_count

              FROM     sys.partitions

              GROUP BY object_id,

                       index_id

              ) pc

        ON     t.object_id              = pc.object_id

           AND ix.index_id              = pc.index_id

WHERE   ps.avg_fragmentation_in_percent > 10

    AND ix.name IS NOT NULL


 

参考: http://blog.miniasp.com/post/2009/01/18/Let-SQL-Server-Tell-You-Which-Indexes-to-Rebuild-or-Reorganize.aspx

posted on 2011-08-05 09:13  中国蓝天  阅读(1026)  评论(0编辑  收藏  举报

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