I have a table which has two partitions (by range): first_half and second_half based on a column "INSERT_DAY".
I need to add sub partitions "SUCCESS" and "NONSUCCESS" based on the values of another column "STATUS" (subpartition by list) i.e. I need to transform my range partition to composite (range-list) partition. I do not wish to drop existing tables or partitions. What is the ALTER query for this?
I have a partition table which is partition on date_loaded column, can i alter the partition key column to orig_date_loaded column
View 9 Replies View Relatedi have in tb1 data like and i need the latest status changed date : when date on which the status of the task changes.for the data below for taskno :
1 i need to get the o/p : 1/2/2013and for taskno :
2 I need the o/p 1/4/2013 taskno :
upd_dateold_status new_status11/1/2013openWIP11/2/2013WIPClosed11/3/2013closed closed21/1/2013openWIP21/2/2013WIPmore_info_needed21/3/2013more_info_neededWIP21/4/2013WIPclosed im writing the query as below select taskno, max(upd_date) MAX_UPD_DATE from (select taskno, upd_date, old_status, new_status, rank from (select taskno, upd_date, old_status, new_status, rank() over (partition by taskno, old_status, new_status order by upd_date asc) as rankfrom tb1)where rank = 1and old_status <> new_status)group by taskno;
IS there any more efficient way to write this query as the clause "old_status <> new_status" is taking a toll on the performance
i have table with range partition and list sub-partition..can i add one more list sub-partition if it is not possible , i have to drop first sub-partition.
View 6 Replies View RelatedWe have some tables in our database in which for loading data we have the setup in place to do Exchange partition after data load into staging. Today we did changes to column length to one pair of main and staging table. Post that Exchange partition stop working.
View 1 Replies View RelatedI Know we can create dynamic partitions on table in oracle 11g. Is it possible to create normal partition and sub partition both dynamically.I have to create Normal partition range on date and sub partition list on Batch ID (varchar).
View 3 Replies View RelatedI have a table that partitioned into six partitions. each partitions placed in different table space and every two table space placed it on a different hardisk
when I will do query select with the non-partition keys condition, how the search process ? whether the sequence (scan sequentially from partition 1 to partition 6) or partition in a hardisk is accessed at the same time with other partition in other hardisk. ( in the image, partition 1,4 accessed at the same time with partition 2,5 and 3,6)
At present we have a non partitioned table.
Can we apply redefinition and create range partition and hash sub partition on it?
I have a question related to partitions and dividing into subpartitions on the existing table.Situation is as follows:
1. we have an inventory table with a list partition on one column sales_desk_id.
2. This table contains millions of records. Due to concurrency and due to high amount of data inserts, now there is a need to make sub partitions based on sale_date.
Question: is there any way to make the subpartitions without dropping the tables?
BANNER
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Oracle Database 11g Enterprise Edition Release 11.1.0.7.0 - 64bit Production
PL/SQL Release 11.1.0.7.0 - Production
CORE 11.1.0.7.0 Production
TNS for IBM/AIX RISC System/6000: Version 11.1.0.7.0 - Production
NLSRTL Version 11.1.0.7.0 - Production
How can we partition a table based on date if it does not have a date column.
Actually I have to compare two tables on daily basis and fetch few rows from those two tables and enter it to a third table.But both these tables does not have a date column.
I am confused if i need to alter those tables and add date column or if there is some way in which i can compare the data from the two tables for that particular day only and not the whole table data.
Can I add range sub partition to a hash partition table. Example like this.
CREATE TABLE test
(
test_id VARCHAR2(10 ) ,
test_TYPE VARCHAR2(5) ,
CREATE_DATE date
)
partition by hash (test_id, test_type)
Partitions 3
SUBPARTITION BY RANGE (CREATE_DATE);
When Tried, I am getting syntax error as invalid option.
I have two tables Activity and Activity1.
Activity Structure
ACTIVITY_TYPE CHAR (1) NOT NULL,
ACTIVITY_DATE DATE DEFAULT sysdate NOT NULL,
ACTIVITY_ON VARCHAR2 (30) NOT NULL,
REFERENCE_NO VARCHAR2 (19),
CHILD_REFERENCE_NO VARCHAR2 (19),
USER_ID VARCHAR2 (30) DEFAULT user NOT NULL,
TERMINAL VARCHAR2 (30) DEFAULT userenv ('TERMINAL') NOT NULL )
Activity1 Structure Which I have Done Partitioning When I insert data from Activity to Activity1 it gives that error ORA-14400: inserted partition key does not map to any partition what I am doing wrong
CREATE TABLE ACTIVITY1(
ACTIVITY_TYPE CHAR (1) NOT NULL,
ACTIVITY_DATE DATE DEFAULT sysdate NOT NULL,
ACTIVITY_ON VARCHAR2 (30) NOT NULL,
REFERENCE_NO VARCHAR2 (19),
[code]....
Insert Statement
insert into ACTIVITY1(ACTIVITY_TYPE,
ACTIVITY_DATE,
ACTIVITY_ON,
REFERENCE_NO,
CHILD_REFERENCE_NO,
[code]....
i want to create a new partition for version 2
existing table is as below
create table test
(
name varchar2(100),
version NUMBER(12)
)
[Code]....
I have partition based table one the basis of year month. And we have 8 local indexes on this table. Every month we have to create a new partition and load data into this partition and the volume of the data is around 14million and the load process is taking long time due to indexes. Is it possible to drop the indexes from particular partition?
View 8 Replies View RelatedI have two tables in which one is partitioned table with the following details.
CREATE TABLE "SCOTT"."TBL_MITTAL"
("ACCOUNT_NAME" VARCHAR2(50 BYTE),
"BILL_NO" VARCHAR2(50 BYTE),
"BILL_DATE" VARCHAR2(50 BYTE),
"CLI" VARCHAR2(50 BYTE),
"ANI" VARCHAR2(50 BYTE),
[code].....
When I am trying to insert record from tbl_mittal into tbl_temp table. I am facing "ORA-14400: inserted partition key does not map to any partition" error
SQL> insert into tbl_temp select * from tbl_mittal;
insert into tbl_temp select * from tbl_mittal
*
ERROR at line 1:
ORA-14400: inserted partition key does not map to any partition
AS tbl_mittal is having hugh number of records so I am providing only few rows from tbl_mittal table as test data.
ACCOUNT_NAMEBILL_NOBILL_DATECLIANICHARGE_START_DATEDURFROM_LOCATIONTO_LOCATIONINVOICE_IDCIRCLE
10000010357423128271095119301-Feb-111723006000931488182328-JAN-11 11.30.54.000000000 AM59.04CHANDIGARHJAIPUR271095119
10000011844187128348720198715-Jun-121409470011825531896615-MAY-12 09.10.36.000000000 AM28.03CHANDIGARHBANTWAL348720198
10000011844187128348720198715-Jun-121409470011825531927115-MAY-12 09.10.41.000000000 AM38.32CHANDIGARHBANTWAL348720198
10000011844187128348720198715-Jun-121409470011825531933015-MAY-12 09.10.46.000000000 AM28.81CHANDIGARHBANTWAL348720198
10000011844187128348720198715-Jun-121409470011825531930215-MAY-12 09.10.53.000000000 AM28.96CHANDIGARHBANTWAL348720198
[code].....
I also tried to upload the same data using sqlldr.
sqlldr log file contents is as follows:
Total logical records skipped: 0
Total logical records read: 1857532
Total logical records rejected: 801092
Total logical records discarded: 37
[code].....
So some sqlldr bad file contents is as follows.
100000118441871,283487201987,15-JUN-12,1723958000,9355115251,10-JUN-12 05.56.05.000000 PM,36.99,CHANDIGARH,AMBALA,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,7520533825,10-JUN-12 05.56.14.000000 PM,44.12,CHANDIGARH,AGRA,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,9356452151,10-JUN-12 05.56.17.000000 PM,116.83,CHANDIGARH,JALANDHAR,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,9331223048,10-JUN-12 05.56.21.000000 PM,28.33,CHANDIGARH,KOLKATA,348720198,,
100000118441871,283487201987,15-JUN-12,1723958000,7827927893,10-JUN-12 05.56.24.000000 PM,3384.33,CHANDIGARH,DELHI,348720198,,
[code].....
needed in stored procedure to achieve this...how to get this..(stored procedure)
table:
studentid,sname,partitonid
901,x,null
902,y,null
903,z,null
904,p,null
905,q,null
906,a,null
907,b,null
908,d,null
909,f,null
910,m,null
For the above data set i need to divide into 5 partittions and need to updated the partitonid with the partition number for each partition set,like the below result set
studentid,sname,partitonid
901,x,p1
902,y,p1
903,z,p2
904,p,p2
905,q,p3
906,a,p3
907,b,p4
908,d,p4
909,f,p5
910,m,p5
write ALTER STATEMENT for adding new partition P1 and SUBPARTITION P1_201001 and P1_201002.
TABLE
=======
CREATE TABLE TEST
(
"REPORT_ID" NUMBER,
"MONTH_ID" NUMBER,
"GROUP_ID" NUMBER,
"AGE_GROUP_ID" NUMBER,
[code]...........
we have our production database running on 11.2.0.2. I have a user table which has partitions. i would like to drop the partition and seeing weird issue:
The command to drop the partition i have is:
alter table SA.accounts drop partition dec_20111203 update global indexes;
Say after 10 minutes, the session is terminated and we are seeing the following: ORA-00028. We don't have a process that kills the session so we are trying to find the root cause why the session is being killed. what might be going on? also as an alternative - i was looking to do the following and want to know - if this works or not:
ALTER TABLE sa.accounts TRUNCATE PARTITION dec_20111203;
Then try to do the drop partition again.
In my 11g Oracle i have the following issue:
create table TEST
(
COL_DATE DATE
)
partition by range (COL_DATE)
(
partition TEST2012 values less than (TO_DATE(' 2013-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
[Code]....
when i execute this i receive the following error:
ORA-02149: Specified partition does not exist?
We are using oracle 10g. with our code, Currently Oracle partitions are size the same way, each partition is using 10MB for data and 12MB for indexes (with the 6 default indexes); even of very few records are written in the partition.
We create partitions in advance as a part of nightly job with 10 minutes duration.Can some intelligence can be added where based on statistics we can decide the size of partition dynamically? Lot of space is getting wasted because of this reason.
How to find the size pf a partition in a partition table?I guess we need to query views like dba_tab_partitions but I am not very sure. will running dbms_stats.gather_table_stats('schema_name,'table_name,'partition_name')
View 3 Replies View RelatedI used the Exchange Partition feature to swap segments between 2 tables- one Partitioned, and one Non-Partitioned. The exchange went well. However, all the data in the partitioned table has gone to the partition which stores the maxbound values.
/** actual table names changed due to client confidentiality issues */
-- Drop the 2 intermediate tables if they already exist
drop table ordered_inv_bkp cascade constraints ;
drop table ordered_inv_t cascade constraints ;
/**
1st create a Non-Partitioned Table from ORDERED_INV and then add the primary key and unique index(s):
*/
create table ordered_inv_bkp as select * from ordered_inv ;
alter table ordered_inv_bkp add constraint ordinvb_pk primary key (ordinv_id) ;
--
create unique index ordinv_scinv_uix on ordered_inv_bkp(
SCP_ID ASC,
[code]....
-- Next, we have to create a partitioned table ORDERED_INV_T with a similar
-- structure as ORDERED_INV.
-- This is a bit tricky, and involves a pl/sql code
declare
l_dt_start DATE;
l_ptn VARCHAR2(50);
cnt PLS_INTEGER;
l_cnt_initial PLS_INTEGER;
ts_name VARCHAR2(50);
l_sql VARCHAR2(10000);
ts_indx VARCHAR2(100);
[code]....
-- Add section to set default values for the intermediate table OL_ORDERED_INV_T
FOR crec_cols IN (
SELECT u.column_name ,u.nullable, u.data_default,u.table_name
FROM USER_TAB_COLUMNS u WHERE
u.table_name ='ORDERED_INV' AND
u.data_default IS NOT NULL )
LOOP
[code]....
-- Next, use exchange partition for actual swipe
-- Between ordered_inv_t and ordered_inv_bkp
-- Analyze both tables : ordered_inv_t and ordered_inv_bkp
BEGIN
DBMS_STATS.GATHER_TABLE_STATS(OWNNAME => 'HENRY220', TABNAME => 'ORDERED_INV_T');
DBMS_STATS.GATHER_TABLE_STATS(OWNNAME => 'HENRY220', TABNAME =>'ORDERED_INV_BKP');
END;
/
SET TIMING ON;
[code]....
OS=Windows 2003
DB = 11gR2
We got a request to take offline for few unused partitions and move it to a another drive (lease used).Please set the following partitions offline in PreProd:
MONTH_20110901
MONTH_20111001
MONTH_20111101
MONTH_20111201
MONTH_20120101
MONTH_20120201
Please move the offline partitions to the E disk.How can we do this ?
I was confused by partitioed table, when i select a partition of table, how does oracle to scan blocks? it scan all blocks of table or scan a single partition blocks only?
SQL> Explain Plan For
2 Select Count(1) From Tb_Hxl_List Partition(p_L3);
Explained.
SQL> Select * From Table(dbms_xplan.display);
PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 18 (0)| 00:00:01 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | PARTITION LIST SINGLE| | 33115 | 18 (0)| 00:00:01 |
| 3 | TABLE ACCESS FULL | TB_HXL_LIST | 33115 | 18 (0)| 00:00:01 |
I'm using Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production and TNS for Linux: Version 11.2.0.3.0 - Production.Requirement is to create a script to add a LIST partition to some selected tables in a schema (tables do not have data, they are not partitioned). There are about 300 such tables (can vary) and their names are maintained in a separate table. Example -Existing table
-CREATE TABLE test_part(id number (2),
name varchar2(20),
audit_userid number (9));
Expected table
-CREATE TABLE test_part(id number (2), name varchar2(20), audit_userid number (9))
PARTITION BY LIST (audit_userid) (PARTITION p1_audit_userid VALUES (1));
Ultimate goal is to add more partitions based on the amount of data to be populated.
I'm trying to do a pivot query in oracle to get the years from a column and make a separate column for each. I found an example of the code to use on the internet and i changed it for my own tables but i'm getting errors. Namely a "FROM keyword not where expected" error at the beginning of the 'avg(...' statements.
I have copied the code used in
select stud_id, 2006, 2007, 2008, 2009
from (
select stud_id,
avg(case when year=2006 then ((present/poss)*100) else null end) 2006,
avg(case when year=2007 then ((present/poss)*100) else null end) 2007,
avg(case when year=2008 then ((present/poss)*100) else null end) 2008,
avg(case when year=2009 then ((present/poss)*100) else null end) 2009
from attendance.vw_all_attendance_perc
group by stud_id
);
I think that performance better partition table than non-partition table. How to assure partition table is better than non-partition table at SELECT operation?
I have compare a specific query EXPLAIN PLAN at partition table and non-partition table. both tables data is same. Is it true way or not?
I have a partitioned table in which at the first level it is been partitioned based on date column(C1) and within in each partition it is again sub-partitioned based on a column(C1) which is a numeric value now at first level(C1 column level) there are 6 partitions and in the last partition i want to add another partition how can i do that.
View 8 Replies View RelatedI am trying to add partition to table without partition
with following code
ALTER TABLE ACC_LOC1_TAB
ADD PARTITION testpart BY RANGE (ALT_AUTHDT)
(PARTITION UPTO_2010 values less than (31-mar-2010),
PARTITION APR_JUN_10 VALUES less than (30-JUN-2010),
PARTITION JUL_SEP_10 VALUES less than (30-SEP-2010),
PARTITION OCT_DEC_10 VALUES less than (31-DEC-2010),
PARTITION JAN_MAR_11 VALUES less than (31-MAR-2011))
it will raise error ora-14020
I am using Oracle 11.2.0.1 Oracle Database.I have a table with 10 Million records and it's a Non Partitioned Table.
1) I would like to partition the table (with partition by range ) without creating new table . I should do it in the existing table itself (not sure DBMS_ REDEFINITION is the only option ) (or) can i use alter table ...?
2) Add one partition which will have data for the unspecified range.