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Sakthivel avatar image
Sakthivel asked

SQL Server 2014 Partition Function

Hi All, I have a backup copy of database from SQL Server 2005 Enterprise Edition and I restored that backup in SQL Server 2014 Standard Edition. Then I got an error. See below: Msg 3013, Level 16, State 1, Line 8 RESTORE DATABASE is terminating abnormally. Msg 905, Level 21, State 1, Line 8 Database 'TestDB' cannot be started in this edition of SQL Server because it contains a partition function 'TestDB_PartitionRange'. Only Enterprise edition of SQL Server supports partitioning. Msg 933, Level 21, State 1, Line 8 Database 'TestDB' cannot be started because some of the database functionality is not available in the current edition of SQL Server. From this error I checked my 2005 database there is no partition scheme so can you suggest any solution for above error? Thanks & Regards, Sakthi
sql serverpartitionenterprise-edition
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sjimmo avatar image
sjimmo answered
You can run this code to check for partitions: select * from sys.partitions You can run this to get a list of partitioned tables: select * from sys.partitions p inner join sys.tables t on p.object_id = t.object_id where p.partition_number <> 1 Here is a thread from SQLServerCentral on finding partitions: http://www.sqlservercentral.com/Forums/Topic992539-391-1.aspx Hope this helps.
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Thanks for your valuable response and also i checked with that database there is no partition table available.Any other way to find partition scheme and table,,,,,,,
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perrywhittle avatar image
perrywhittle answered
check sys.partition_schemes for an existing scheme, the error suggests you have a portioned table in the database
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Thanks for your valuable response and also i checked with that database there is no partition table available.Any other way to find partition scheme and table,,,,,,,
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ThomasRushton answered
Given that the error message makes reference to a partition function, have you tried looking in sys.partition_functions? It is, after all, possible to have a partition function without associating it with a partition scheme.
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