SQL & PL/SQL :: Use Of Local And Global Partition Index
Aug 17, 2010What is the use of Local and Global Partition Index?
View 1 RepliesWhat is the use of Local and Global Partition Index?
View 1 RepliesI'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;
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 RelatedI have a global index and I want to convert it to local index.Is there a way to recreate local index with out dropping the global index.
I can create a local index first and then drop the global index. But is there a way to create it with out dropping the global index, just convert it.
Version Info: 11.2 on Solaris 10
I have a partitioned table like below. I want to create a B-Tree index on SALES_RGN column which is neither the part of Primary key or the Partitioned key. Should I create this index as local or Global ?
CREATE TABLE sales_dtl
(
txn_id number (9),
salesman_id number(5),
salesman_name varchar2(30),
sales_rgn varchar2(10), -----------------------------> This column needs to be indexed
sales_amount number(10),
sales_date date,
constraint pk_sales_dtl primary key (txn_id)
[code]....
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]...
RDMS Version : 11.2.0.2
Platfomrm : AIX 6.1
For partitioned tables (RANGE, LIST types ) what are circumstances when a Global or a Local Index become UNUSABLE ? I was told that in some circumstances where the Indexes become UNUSABLE , not even a SELECT query against the table will work. Is this true ? For partitioned tables with Global Index, the global index will be listed in DBA_INDEXES . Right ?
1)What is the use of inventory?
2) Global inventory is found by checking the oraInst.loc file. How to find the location of Local inventory?
3)Why is every patchset(11.2.0.2) now a full release?
getting how many local and global indexes on particular oracle table
View 2 Replies View RelatedI 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 created a table below, my TL asked me to create a local unique constraint for the below table.
I went through all sites and could not find the correct solution, how to create LOCAL UNIQUE CONSTRAINT ON SUB PARTITION TABLE and LOCAL UNIQUE INDEX ON PARTITION TABLE. Creating Local Unique constraint should take care of creating local unique index creation.
Unique key columns are DET,GDS,ARRIVE_DT
CREATE TABLE SUB_PAR_TAB
(
ID VARCHAR2(100) NOT NULL,
REGION VARCHAR2(40) NOT NULL,
SOURCE VARCHAR2(80) NOT NULL,
DET VARCHAR2(80) NOT NULL,
GDS VARCHAR2(40) NOT NULL,
ARRIVE_DTDATE,
SYS_SOURCE VARCHAR2(25) ,
[code]........
if i have local index, does query which uses this index always have to scan all partitions?
For example, I have hash-partitioned table on column A. There is local index on column B. I am querying:
select * from my_table where A='ABC' and B>123.
I know that records with A='ABC' are in 2nd partition. Does only 2nd partition of local index will be scanned or all of them must be scanned?
I am trying to use Global temporary tables, and index on this table to get my results faster. I can see even if I run any query on this table, it does full table scan and not Index scan..
create global temporary table abc_tab on commit preserve rows
as select a,b,c from xyz;
create index lmn on abc_tab(a,b,c)
I Have created a interval partitioned table with local index.But when automatic partitions are created based record insertion, whether local indexes will be created for each newly created partition or not? If created, how to check
Below is the code which I tried
CREATE TABLE interval_date
(
date1 date,
days VARCHAR2(50)
)
PARTITION BY RANGE(date1)
INTERVAL (NUMTODSINTERVAL(45,'DAY'))
[code]....
2day i was dropping few unwanted index from the data base, By mistake i removed the local partitioned index , So i want to recreate that index.i create the index, will the partitioned index updates when we add partitioned to the tables.
View 2 Replies View RelatedI have a partioned (by row_create_date) table, lets called it TABLE_X, which has about 300 million records. This table has 7 columns including the primary key and a non-unique, locally partitioned column called trace_id; 99% of queries access this table via this column.
Lately, querying TABLE_X via the trace_id has been very very bad. Queries run for over 1 hr in some cases. So we decided to change the index for trace_id to a global index. Now queries against TABLE_X return in seconds. So far so good.
However, when the query has to join TABLE_X to another table, the query sometimes runs for over 1 hours; back to the same old problem. Here is an illustration;
SELECT COUNT(1) FROM TABLE_X WHERE TRACE_ID = 'XXXXX';
-- returns in seconds
SELECT COUNT(1)
FROM TABLE_X,
TABLE_Y
WHERE TABLE_X.TRACE_ID = 'XXXXX'
AND TABLE_X.TRACE_D = TABLE_Y.TRACE_ID;
-- runs for over 1 hr, even when TABLE_Y.TRACE_ID is a primary key.
I have table with 4 partition by range partition. I am loading the table in bulk mode to latest partition. Before I load , I dropped the index and after Load I will be creating index. So when I am dropping index, it is dropping index from all the partitions and when creating the index, I am creating the index for all partitions. When I am creating index using local, it is telling you have to create local index for all partitions at the same time. because of that I have to drop and recreate all indexes again. Again I have to gather stats for whole table .
I was thinking we can build index for one partition and index should remain as is for old partitions If this is not the case, how do I plan my load for a partitioned table using bulk mode to latest partition.
I have a table which if approc 5GB in size.I am trying to create a partition table from it. I created the new table,but when i am creating local indexes, i am getting an error when i create a unique key. I ll post the creation scripts:
CREATE TABLE A_PT_ORDREG
(
BUS_UNIT VARCHAR2(5) NOT NULL,
ORDER_NO VARCHAR2(15) NOT NULL,
PT_CODE VARCHAR2(15) NOT NULL,
REG_NO VARCHAR2(15) NOT NULL,
AS_APPLICATION VARCHAR2(5) NOT NULL,
[code].....
THE INDEX:
CREATE UNIQUE INDEX PK_A_PT_ORDREG ON A_PT_ORDREG (BUS_UNIT, ORDER_NO, SEQ_NO) local;
I am getting the following error:
ORA-14039: partitioning columns must form a subset of key columns of a UNIQUE index
how data is distributed on a local non prefixed index.Let's say I have a table calles SALES and it holds data for the last year partitioned by a date column called month (So that's 12 partitions, one for each month).Now i'm creating the following index: CREATE INDEX TESTIDX ON SALES(salesman_id) local; What will happen is that I'll have 12 partitions on the index but to which partitions in the table are they referencing ?
View 1 Replies View Relatedhow 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?
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 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 RelatedWe 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?
SNRCTL> stat LISTENER
Connecting to (DESCRIPTION=(ADDRESS=(PROTOCOL=IPC)(KEY=LISTENER)))
STATUS of the LISTENER
------------------------
Alias LISTENER
Version TNSLSNR for IBM/AIX RISC System/6000: Version 11.2.0.1.0 - Production
Start Date 19-JAN-2013 00:50:10
Uptime 0 days 0 hr. 29 min. 51 sec
Trace Level off
Security ON: Local OS Authentication
[code]....
In every oracle documentation for e.g:11.2 Scan and Node TNS Listener Setup Examples [ID 1070607.1] we found the local listener status showing both local-ip and vip. Why is not showing in our case?