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Query SQL in compatibiliy level 80 error

Posted on 2016-09-02
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Last Modified: 2016-09-02
Good afternoon,

I need to execute the query below to find out the use of a table, but the base is on this basis compatibility level is 80, the error.

Does anyone know how I can adapt this script to not introduce error?

Msg 102, Level 15, State 1, Line 9
Incorrect syntax near ')'.

Queries that I need to run are attached

Thank you
query1.txt
query2.txt
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Comment
Question by:Alex
3 Comments
 
LVL 29

Expert Comment

by:Olaf Doschke
ID: 41782257
The solution is, you can use DB_ID(), but not directly inside a call as parameter. See for example here: http://www.sqlservercentral.com/Forums/Topic992205-391-1.aspx

If you depend on compatibility levels, because the database moved recently and you don't want to risc a compatibility issue, fix this as soon as possible, testing all your sql code to see whether the database really needs this limitation. You never do yourself a favor in keeping anything in some compatibility level, as it hinders you using new things, you don't improve.

Bye, Olaf.
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Accepted Solution

by:
Scott Pletcher earned 500 total points
ID: 41782318
Query 1, as below.  Do similar thing for query 2.

DECLARE @db_id smallint
SET @db_id = DB_ID()

SELECT o.name AS [Table_Name], x.name AS [Index_Name],
       i.partition_number AS [Partition],
       i.index_id AS [Index_ID], x.type_desc AS [Index_Type],
       i.leaf_update_count * 100.0 /
           (i.range_scan_count + i.leaf_insert_count
            + i.leaf_delete_count + i.leaf_update_count
            + i.leaf_page_merge_count + i.singleton_lookup_count
           ) AS [Percent_Update]
FROM sys.dm_db_index_operational_stats (@db_id, NULL, NULL, NULL) i
JOIN sys.objects o ON o.object_id = i.object_id
JOIN sys.indexes x ON x.object_id = i.object_id AND x.index_id = i.index_id
WHERE (i.range_scan_count + i.leaf_insert_count
       + i.leaf_delete_count + leaf_update_count
       + i.leaf_page_merge_count + i.singleton_lookup_count) != 0
AND objectproperty(i.object_id,'IsUserTable') = 1
ORDER BY [Percent_Update] ASC
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Author Closing Comment

by:Alex
ID: 41782401
thank you
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