SQL & PL/SQL :: Finding The Corresponding Index Partition?
Aug 1, 2012I have a partitioned tables and local partitioned index on the same.
I want to know that a particular index partition belong to which table partition.
I have a partitioned tables and local partitioned index on the same.
I want to know that a particular index partition belong to which table partition.
I am facing the error "ORA-01502: index or partition of such index is in unusable state " while loading the text data using
sql loader with direct path (direct = Y ,rows = 10000) option. Table consists an composite non unique index. If I query the dba indexes for the effected index it shows the index status as VALID. There was no maintaince done on the effected table or index. I have tried loading the same data using conventional path but didn't found any issues for the same.
I have read and used the AWR script (mentioned in the page Finding unused index for finding unused customised (Z) indexes in our SAP system using oracle 10.2.0.2 as the SAP database.
But this returns no rows. Is there any precondition? I want to know how much / many times the indexes are used...We are smelling that lot of unused index are there in the database.
how to partition and index my table for a special problem.
The table:
CREATE TABLE TEST (
ID NUMBER PRIMARY KEY,
U_VALUE NUMBER NOT NULL, -- Ranges from 0 - 30.000.000
O_VALUE NUMBER NOT NULL, -- Ranges from U_VALUE - 30.000.000
CREATE_TS TIMESTAMP DEFAULT SYSTIMESTAMP NOT NULL,
UPDATE_TS TIMESTAMP NOT NULL,
ITEM_TYPE NUMBER NOT NULL --<< Only 4 different values >>
);
As you can see, U_VALUE is ALWAYS lower than O_VALUE.I need to have the CREATE_TS in either main- or subpartition do drop the partitions after some time so i don,t have to use DELETE statements. The table has 360 millions rows.
The application has only 8 query which will always use a WHERE clause like this:
SELECT * FROM TEST
WHERE U_VALUE <= :1 AND O_VALUE => :2 AND ITEM_TYPE = :3
1. Is there any good technique how to create a good index for the queries (application will execute 120 queries per second)?
2. how to partition this table?
What is the use of Local and Global Partition Index?
View 1 Replies View RelatedI have an IOT table with partitioned on list. I have p1,p2 and p3 partitions. Now I would like to create a bitmap index on partition key.
I gave ALTER TABLE .. MOVE MAPPING TABLE;
But getting the below error,
ORA-28660: Partitioned Index-Organized table may not be Moved as a whole.
I am trying to disable the index i am getting the error like
ORA-02243: invalid ALTER INDEX or ALTER MATERIALIZED VIEW option
the table have partition. how to disable the index in partition table.
Index with following PARTITIONS. Index rebuild is extremely slow. Below 2 Alter index ..rebuild... took 10 hours to complete. Because of this queries which based on this index are really slow.
SYS@DB AS SYSDBA> select partition_name,tablespace_name,bytes/1024 KB from dba_segments where segment_name='KSTN';
PARTITION_NAME TABLESPACE_NAME KB
------------------------------ ------------------------------ ----------
REB_IDX_1 TS_REB 64
REB_IDX_2 TS_REB 64
REB_IDX_3 INDEX 64
REB_IDX_4 TS_REB 64
[code]....
Is there any rule in documentation, when create partition index? For tables, it is said to partition when table is greater than 2GB, but what about index? WHat size it should have to partition?
View 1 Replies View Relatedafter merging two partitions into single partition (partition is by list) of a table ,when i analyzed the table it is giving this error : ORA-01502 INDEX TEST.PK_ID or partition of such index is in unusable state.
View 7 Replies View Relatedora 20000 index partition in an unusable state. what can i do
View 6 Replies View RelatedI'm trying to split a table partition.
whether the below syntax are correct.
If local index used
ALTER TABLE SNYT.PART_ESTD
SPLIT PARTITION ESTD_M13_S22 AT ('ESTD', 13, '22')
INTO (PARTITION ESTD_M13_S21, PARTITION ESTD_M13_S22)
update indexes;
(per http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:1401247200346349807)
=================================================================
If Global index used
ALTER TABLE SNYT.PART_ESTD
SPLIT PARTITION ESTD_M13_S22 AT ('ESTD', 13, '22')
INTO (PARTITION ESTD_M13_S21, PARTITION ESTD_M13_S22)
UPDATE GLOBAL INDEXES;
We have a large customer table so first thought was to partition.Also we see two union alls in the plan - can we introduce parallelism? Below is the plan - have attached a text file if difficult to read
SELECT V_IDENTIFIER_LOOKUP.UID_V_IDENTIFIER_LOOKUP AS "UID",
V_IDENTIFIER_LOOKUP.ABA, V_IDENTIFIER_LOOKUP.ADDRESS1,
V_IDENTIFIER_LOOKUP.ADDRESS2, V_IDENTIFIER_LOOKUP.ADDRESS3,
V_IDENTIFIER_LOOKUP.ADDRESS4, V_IDENTIFIER_LOOKUP.ALIAS,
V_IDENTIFIER_LOOKUP.CITY, V_IDENTIFIER_LOOKUP.COUNTRYCODE,
V_IDENTIFIER_LOOKUP.CUST_CODE, V_IDENTIFIER_LOOKUP.CUST_NAME,
V_IDENTIFIER_LOOKUP.HEAD_OFFICE_IN,
V_IDENTIFIER_LOOKUP.IDENTIFIER,
V_IDENTIFIER_LOOKUP.IDENTIFIER_TYPE,
[code]...
Is it possible to build index partition in parallel?I tried following command
alter index <index_name> rebuild partition <partition name> online parallel 5;
It executed without complaining, but want to know if index partitions can be build in parallel?
I am rebuilding some UNUSABLE local index partitions on Oracle 8.1.7.4.0 (64bit) database . The platform is a HPUX machine.
The DDL of the partition table/indexes:
=========================
CREATE TABLE TESTME
( INST_NO CHAR(3) NOT NULL,
ACCT_NO CHAR(16) NOT NULL,
REC_NO CHAR(9) NOT NULL,
TRAN_TYPE CHAR(2) DEFAULT ' ',
STAT CHAR(2) DEFAULT ' ',
[code]...
splitting a table partition without making its primary key index ar any other indexes unusable.
I think it is possible to do so 10g onwards.
DB Details:
Oracle RDBMS 11.2.0.3, HP-Ux B.11.31, OLTP
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 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?
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]...........
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 am on 11.2.0.3 Enterprise Edition. We are using the new feature "Composite Domain Index" for a Domain index on a very large table (>250.000.000 rows). It really works with mixed queries. We added two number columns using FILTER BY.We have lots of DML on this table. Therefore, we are executing synchronize and optimize once the week. The synch behaves pretty normal. But "optimize_index" takes a very very long time to complete. I have switsched on 'logging' for the optimize process. The $I table takes some time but is finished normally. But the optimization of the $S table (that is the table created for the CDI feature) is running over 12 hours now - and far from being finished. From the logfile, I can see that it optimizes 1000 rows every 20 minutes. Here is the output of the logfile:
Oracle Text, 11.2.0.3.0
14:33:05 06/26/12 begin logging
14:33:05 06/26/12 event
14:33:05 06/26/12 process $N for optimize: SEQDEV.GEN_GES_DESCRIPTION_CTX_I
14:33:16 06/26/12
14:33:16 06/26/12
[code]....
I haven't found a recommendation from Oracle not to use "optimize_index" for Domain Indexes with CDI. But in my case, it would be much faster just to drop and recreate the Domain Index in question.
I have a huge table (about 60 gb) partition over range. The index on this table is global index created on 4 columns together. I have a query which is running very slowly. The explain plan is showing the use of this global index.Explain plan is not showing pstart and pend because the index is global.
View 6 Replies View Related