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?
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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 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.
Does truncating partition require rebuilding local index partition? (Oracle 10g2) When to use update indexes and when update global indexes?
View 4 Replies View RelatedSo our situation is pretty simple. We have 3 tables.
A, B and C
the model is A->>B->>C
Currently A, B and C are range partitioned on a key created_date however it's typical that only C is every qualfied with created date. There is a foreign key from B -> A and C -> Bhave many queries where the data is identified by state that is indexed currently non partitioned on columns in A ... there are also indexes on the foreign keys that get from C -> B -> A. Again these are non partitioned indexes at this time.
It is typical that we qualifier A on either account or user or both. There are indexes (non partitioned on these) We have a problem with now because many of the queries use leading wildcards ie. account like '%ACCOUNT' etc. This often results in large full table scans. Our solution has been to remove the leading wildcard.
We are wondering how we can benefit from partitioning and or sub partitioning table A. since it's partitioned on created_date but rarely qualified by that. We are also wondering where and how we can benefit from either global partitioned index or local partitioned indexes on tables A. We suspect that the index on the foreign key from C to B could be a local partitioned index.
I exchange a partition with into a normal table with UPDATE INDEXES,but i found the index of both table are marked UNUSABLE?
SQL> Create Table tb_hxl_list_part
2 (
3 statedate Number,
4 provcode Number
5 )
6 Partition By List(provcode)
7 (
8 Partition p_1 Values(1)
9 );
Table created.
SQL> Create Unique Index idx_tb_hxl_list_part On tb_hxl_list_part(provcode) Local;
Index created.
SQL> Insert Into tb_hxl_list_part Values(20111101,1);
1 row created.
SQL> commit;
Commit complete.
SQL> Select status From dba_ind_partitions aa
2 Where aa.index_name = 'IDX_TB_HXL_LIST_PART';
STATUS
--------
USABLE
SQL> Create Table tb_hxl_list_part_bak
2 (
3 statedate Number,
4 provcode Number
5 );
Table created.
SQL> Create Unique Index idx_hxl_list_part_bak On tb_hxl_list_part_bak(provcode);
Index created.
SQL> Select status From dba_indexes bb
2 Where bb.index_name = 'IDX_HXL_LIST_PART_BAK';
STATUS
--------
VALID
SQL> Alter Table tb_hxl_list_part
2 Exchange Partition p_1
3 With Table TB_HXL_LIST_PART_bak UPDATE Indexes;
Table altered.
SQL> Select status From dba_ind_partitions aa
2 Where aa.index_name = 'IDX_TB_HXL_LIST_PART';
STATUS
--------
UNUSABLE
SQL> Select status From dba_indexes bb
2 Where bb.index_name = 'IDX_HXL_LIST_PART_BAK';
STATUS
--------
UNUSABLE
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 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 range date partition table month wise which i want to truncate the partition and load for each month data i get from source.
How can i get the partion name based on data value.
e.g
source data
SALES_IDSALES_DT
101/02/2008
101/03/2008
101/04/2008
101/05/2008
target table has same structure as source but has more data but i want to delete only those which is coming from source if it is there in target and then load. so i want to get partition name so that i can truncate and load them.
How to truncate the partition based on the given input. I've below table partitioned on dt field. How can i truncate the whole partition which contain the date 2013/04/30.
CREATE TABLE TST_SUMMARY
(
CUST_ID NUMBER(38) NOT NULL ,
[Code]....
We have fact tables partitioned based on week.
We receive point of sale data on daily basis .
Some times we will receive data for previous week also.
In current system , we are analyzing only one partition(current partition).
Now , we want to analyze all old partitions also based on incoming old trading date.
how can i approach to find partitions for old data (trading_date s)
i am trying to create Procedure which will create the partitions based on the other table date values one per each day.
CREATE OR REPLACE PROCEDURE PARTITION_TEST(PART_DATE_TABLE IN VARCHAR2, TABLE_NAME IN VARCHAR2,SCHEMA_NAME IN VARCHAR2)
AS
V_PART_NM VARCHAR2(20);
V_PART_CNT NUMBER;
V_DATE DATE;
V_SCHEMA_NAME VARCHAR(15);
[Code]..
It is not creating the partitions and even not giving any errors.
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 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]...........
I need to partition a existing table based on varchar2 field (which is actaully date value but storing as character in the table). Using below statement for creating table, but getting error.
create table TST_CUST_ARC
(
interact_id NUMBER(10),
extrn_id VARCHAR2(38),
src_cd VARCHAR2(25),
full_nm VARCHAR2(45),
run_control_date VARCHAR2(13)
[code]....
Getting error : ORA-00907: missing right parenthesis
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.
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 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.
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 alter the table to create partition on non partition table, i have tried and could not create it. Do we have some other means to do it as this is the live table and cannot drop them else will lose the data.
View 1 Replies View Related
