Home » RDBMS Server » Server Administration » how to split a default subpartition (11.1.0.7 windows xp)
how to split a default subpartition [message #526683] |
Wed, 12 October 2011 07:56 |
andy huang
Messages: 498 Registered: July 2011
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Senior Member |
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Dear all,
I create a range-list partition table using the subpartition template,and there is a default subpartition p_default,when i insert a row which provcode values is 6 into the table,and it belong to the default subpartition,how can i spilt it into a new subpartition such as p_6, and make the row belong to the subpartition p_6?
create table tb_hxl_user_split
(
statedate date not null,
provcode number not null,
usernumber varchar2(13) not null,
rem1 varchar2(1024),
create_date date,
create_by number,
last_update_date date,
last_update_by number
)
partition by range (statedate) subpartition by list(provcode)
subpartition template (
subpartition p_1 values (1) ,
subpartition p_2 values (2) ,
subpartition p_3 values (3) ,
subpartition p_4 values (4) ,
subpartition p_5 values (5) ,
subpartition p_default values (default) ,
)
(
partition hxl_user_20111001 values less than
(to_date(' 2011-10-02 00:00:00', 'syyyy-mm-dd hh24:mi:ss', 'nls_calendar=gregorian'))
);
insert into tb_hxl_user_split(statedate,provcode,usernumber) values(date'2011-01-01',1,'12301');
insert into tb_hxl_user_split(statedate,provcode,usernumber) values(date'2011-01-01',2,'12302');
insert into tb_hxl_user_split(statedate,provcode,usernumber) values(date'2011-01-01',3,'12303');
insert into tb_hxl_user_split(statedate,provcode,usernumber) values(date'2011-01-01',4,'12304');
insert into tb_hxl_user_split(statedate,provcode,usernumber) values(date'2011-01-01',5,'12305');
insert into tb_hxl_user_split(statedate,provcode,usernumber) values(date'2011-01-01',6,'12306');
[Updated on: Wed, 12 October 2011 08:24] by Moderator Report message to a moderator
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