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Sql Server 2008 r2 index defrag degree more than x%

Posted on 2013-01-24
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Last Modified: 2013-02-05
Hi,

I´m using a sql server 2008 r2 with more than 2500 tables.

How  can I check all of them and obtain how many of them have more than xx% degree fragmented?

For example tables with more than 75%  index fragmentation degree

Any idea?

regards
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Question by:heze54
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Steve Wales earned 500 total points
ID: 38816622
The following will show you:

SELECT ps.database_id, ps.OBJECT_ID,
ps.index_id, b.name,
ps.avg_fragmentation_in_percent, b.name, ps.page_count
FROM sys.dm_db_index_physical_stats (DB_ID(), NULL, NULL, NULL, NULL) AS ps
INNER JOIN sys.indexes AS b ON ps.OBJECT_ID = b.OBJECT_ID
AND ps.index_id = b.index_id
WHERE ps.database_id = DB_ID()
and ps.avg_fragmentation_in_percent > xx
and ps.page_count > yy
ORDER BY b.name
GO

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Replace xx with your percentage break limit and yy with the minimum number of pages on the index.
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Author Comment

by:heze54
ID: 38817830
Hi,

Could you explain better yy parameters?


Regards
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LVL 22

Expert Comment

by:Steve Wales
ID: 38818906
Page Count is the number of pages in the index.

Usually, it's not worth trying to defrag indexes with only a couple of hundred pages in them, because there is just too few pages in order for it to make a difference.

Play with the parameter in your system and see what you get.  Start with 100 or 200 and see what it shows you.
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Author Comment

by:heze54
ID: 38825954
Hi,


Sintax error?

 SQL Server Database Error: Sintaxis incorrecta cerca de ')'. Line 4
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LVL 22

Expert Comment

by:Steve Wales
ID: 38826670
I'd bet that your SQL Server compatibility mode is set to 80 and it's complaining about the DB_ID() function.

You can get around this by doing a select DB_ID(), find out the value that returns the replace DB_ID() with that integer value on lines 4 and 7.
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Author Closing Comment

by:heze54
ID: 38856041
A++
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