partitions for Exchange works only (no movement records to all partitions)
The use of partitions of exchange to move data from a table that is not partitioned to partitioned table does not move the records of all partitions...
for example
Not partitioned table
create table tab1 as select object_id from user_objects;
partitioned table
create table parttab (OBJECT_ID number) partition by hash (object_id)
(partition p1, partition p2, p3, p4 partition partition);
Exchange
ALTER table parttab Exchange partition p1 with table tab1 without validation;
ALTER table parttab Exchange partition p2 with table tab1 without validation;
ALTER table parttab Exchange partition p3 with table tab1 without validation;
ALTER table parttab Exchange partition p4 with table tab1 without validation;
Select count (1), "P1" as partnam parttab partition (p1)
Union
Select count (1), "P2" as partnam parttab partition (p2)
Union
Select count (1), 'P3' as partnam parttab partition (p3)
Union
Select count (1), 'P4' as partnam parttab partition (p4)
order of partnam;
only P1 is to have the records, other partitions are zero records
Please suggest
Thank you
S
Thank you Kiss. but now the table will have double right recordings... ?
y at - it an otherway?
Another way to do what? Load this HASH partitioned table? Yes - simply INSERT in it as PaulHorth suggested.
What you're trying to load a table partitioned hash isn't sensible. These tables use a HASH algorithm to fill several partitions so that each partition will have the same number of lines.
An EXCHANGE puts ALL the data in a partition; This is just an update of data dictionary.
or I try with just 1 hash partition at the start then swap the records and split... but the split hash partition is possible? NO.
No - you can't divide a HASH partition. See the VLDB and partitioning Doc
http://docs.Oracle.com/CD/B28359_01/server.111/b32024/part_admin.htm#g1010774
Also the use of "without validation" will cause problems for your first example because partition P1 will be given which actually belongs to P2, P3 and P4 partitions.
So if you now create a local unique index on this table, you will be able to use this index to find any data belonging to these other three partitions because Oracle will use on the index partition pruning to determine which partitions to check. And the index for other partitions partitions will have all the data in them all the data being located on the P1 partition.
Try this code example to see what I mean:
drop table myTab cascade constraints;
drop table myTabPar cascade constraints;
drop table myTabPar1 cascade constraints;
create the table myTab (myId number);
create the table MyTabPar (myId number) partition by hash (myId)
(partition p1, partition p2);create the table MyTabPar1 (myId number) partition by hash (myId)
(partition p1, partition p2);create an index on mytabpar (myId) single mytabpar_ndx local
create an index only mytabpar1_ndx on local mytabpar1 (myId)
insert into values myTab (1);
insert into myTab values (2);
insert into mytabpar1 select * from mytab
ALTER table mytabpar Exchange partition p1 with table mytab without validation
commit;
Select * from mytabpar;
MYID
1
2
Select * partition mytabpar (p1)
MYID
1
2
Select * partition mytabpar (p2)
-No data
Select * from mytabpar where myid = 2
MYID
2
Select * from mytabpar where myid = 1
-No data
PLAN_TABLE_OUTPUT
SQL_ID, 1s1gpvm7pf05k, number of children 0
-------------------------------------
Select * from mytabpar where myid = 1
Hash value of plan: 1899509776
-------------------------------------------------------------------------------------------
| ID | Operation | Name | Lines | Bytes | Cost (% CPU). Pstart. Pstop |
-------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | | | 1 (100) | | |
| 1. UNIQUE HASH PARTITION | | 1. 13. 0 (0) | 2. 2.
|* 2 | INDEX UNIQUE SCAN | MYTABPAR_NDX | 1. 13. 0 (0) | 2. 2.
-------------------------------------------------------------------------------------------
Information of predicates (identified by the operation identity card):
---------------------------------------------------
2 - access ("MYID" = 1)
The plan shows Oracle uses the unique index to find the data from a hash of the value ('myid = 1') indicates that only the partition of the local index for partition 2 may possibly have the necessary data.
Of course, it is wrong because the clause "without validation" has been used to put all the data in the first partition.
The typical use case if HASH is involved is when there are secondary hash partitions. Then, you can create a table 'temp' that is partitioned by hash and switch with the partition - you exchange them with the subparts of hash. This is the ONLY use case where EXCHANGE to operate between two partitioned tables: a table is partitioned and sous-partitionnée and the 'temp' table is partitioned but its partitioning scheme is identical to the "sub" mΘthodes of recommandΘes of the other table partitioning.
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