RE: Paritioning Challenge: alternate unique constraint

From: <oracle-l-bounce_at_freelists.org>
Date: Wed, 10 Mar 2004 16:48:27 -0800
Message-ID: <B5C5F99D765BB744B54FFDF35F60262119FBE2@irvmbxw02>

Could you expand on this please Mr. Cave?

You said "If you did have a number of local indexes, Oracle would have to scan each index before it inserted a new row in any partition, which would likely be a rather poorly performing option."

I'm not sure what this means. In my example below I have a table hash partitioned by column A, with unique index 1 global range partitioned by column B, and unique index 2 global range partitioned by column C. Are you saying that the uniqueness for columns B and C can be enforced by a better algorithm because indexes 1 and 2 are global, rather than local?

SQL> create table t (n1 number, n2 number, n3 number) 2 partition by hash (n1) partitions 2 ; Table créée.

SQL> create unique index tgui1 on t (n2) global partition by range (n2) 2 (partition values less than (100), partition values less than (maxvalue)) ; Index créé.

SQL> create unique index tgui2 on t (n3) global partition by range (n3) 2 (partition values less than (100), partition values less than (maxvalue)) ; Index créé.

> -----Original Message-----
> Justin Cave (DDBC)
>
> As I understand it, you want to create local indexes on a
> partitioned table that do not include the partition key.
>
> Logically, this sort of construct doesn't strike me as
> possible. Since uniqueness has to apply to the whole table,
> you logically need to, in this case, have a single object to
> store all possible first & last names. This would require a
> global index. If you did have a number of local indexes,
> Oracle would have to scan each index before it inserted a new
> row in any partition, which would likely be a rather poorly
> performing option.

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Received on Wed Mar 10 2004 - 18:53:49 CST