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;
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?
I have one user CD_APP. I have one partition table CD.T_FCDR_DT. User has got ALTER/INSERT/UPDATE/DELETE/SELECT privileges on the table..
Now when I try to drop a partition, I get error as below: ------------------------------------- SQL> show user USER is "CD_APP" SQL> ALTER TABLE CD.T_FCDR_DT DROP PARTITION D01 UPDATE GLOBAL INDEXES; ALTER TABLE CD.T_FCDR_DT DROP PARTITION D01 UPDATE GLOBAL INDEXES * ERROR at line 1: ORA-01031: insufficient privileges -------------------------------------- Do I have to grant some other privileges for this user.
I am Using Oracle 11.2.0.3. I had a script that droped partitions for a table which should have dropped nearly 30 gb of data. We have data and index tablespace.
Query of dba_free_space showed no change on data tablespace. However, index tablespace showed increase in free space.
select sum(bytes / (1024 * 1024 * 1024)) "Size (GB)" from dba_free_space where tablespace_name ='&tbs_name'
I am using following command to drop partition
ALTER TABLE table_name DROP PARTITION "partition_name" UPDATE GLOBAL INDEXES;
What should be done to increase free space shown for tablespace after partition drop?
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.
I 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).
Through an Oracle Apex application I need to create/drop a user/schema in another Oracle database. i.e., create/drop user remotely using an Oracle Apex application.
I 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)
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 -------------------------------------------------------------------------------- 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
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.
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,
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.
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
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:
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?
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')
I 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
-- 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;
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:
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);
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 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.
I 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))
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.
