Partitioned index
Using Oracle 11.2.0.3We are evalauating partitiong stragetegies with a view to the realization of gains from perfomnace in reports in particular.
How effeicient are partitioned indexes in this by example anti-terrorism just partitioned index aan table is not partitioned.
One big fact to durrogate keys that have bitmpa indxese table which link to accentuate associated key dimensions.
Patitioning given bitmap index which links to the largest dimension and partitiong Kay dimesnion dimension laregts.
Reflections on partitioned indexes would be particularly useful.
Thank you
user5716448 wrote:
Using Oracle 11.2.0.3We are evalauating partitiong stragetegies with a view to the realization of gains from perfomnace in reports in particular.
How effeicient are partitioned indexes in this by example anti-terrorism just partitioned index aan table is not partitioned.
One big fact to durrogate keys that have bitmpa indxese table which link to accentuate associated key dimensions.
Patitioning given bitmap index which links to the largest dimension and partitiong Kay dimesnion dimension laregts.
Reflections on partitioned indexes would be particularly useful.
Thank you
It is not possible to create a partitioned index bitmp on a non-partitioned table. Bitmap indexes can be local partitioned only.
--
John Watson
Oracle Certified Master s/n
http://skillbuilders.com
Tags: Database
Similar Questions
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deletion of a partitioned index
Hi friends,
I use 10.2.0.4 oracle on solaris.
I have several partitioned index with the 2011 created on a daily basis. I tried to drop one of the indexes and got the below error.
SQL > ALTER INDEX QOSDEV. PK_RATE_CISCOMEMORYPOOL DROP PARTITION 'OCTOBER 5, 2012 '.
ALTER INDEX QOSDEV. PK_RATE_CISMEPOOL DROP PARTITION 'OCTOBER 5, 2012 '.
Error on line 2
ORA-14076: submitted alter index partition/subpartition operation is not valid for local partitioned indexes
Script done on line 2.
I ask you how to remove these partitions.
Thank you
DBApps
Hello
Try-
ALTER drop partition table RATE_CISCOMEMORYPOOL 'October 5, 2012;
Anand
-
Hi all
I created 2 local partitioned index. The indexes are indexes of function. the table size is 2.3 T.
I created the first clue that it took 14 hours to create, and even to analyze and it works fine now. Then, I created second index. But now it does not.
What should I do?
version 11.1.0.6
RAC, ASM
Thanks in advance
Published by: disaster on April 10, 2011 21:43
Published by: disaster on April 10, 2011 21:51Dear Sir
---------------------------------------------------------------------------------------------------------- | Id | Operation | Name | Starts | E-Rows | A-Rows | A-Time | Buffers | Reads | ---------------------------------------------------------------------------------------------------------- | 1 | PARTITION RANGE SINGLE| | 1 | 1 | 82 |00:03:37.92 | 534K| 534K| |* 2 | TABLE ACCESS FULL | TBL | 1 | 1 | 82 |00:03:37.92 | 534K| 534K| ----------------------------------------------------------------------------------------------------------
It is the plan of the real explanation followed by the SQL engine to run your query
The Oracle optimizer is the estimate (based on the statistics that you have collected about index and table) that your query will return only 1 rank (E-lines = 1) while in reality (when the query has been executed) it's return of 82 lines (A-Rows = 82) within 3 minutes and 37 seconds (A-Time = 00:03:37.92)
It is clear that your index function that is not used.
You should be aware that when you create a function based index, oracle will create a virtual column that is hidden behind the scene.
Try to gather statistics on this column using dbms_stats.
Please, try first to TEST
BEGIN DBMS_STATS.gather_table_stats (ownname => user, tabname => 'TBL', CASCADE => TRUE, method_opt => 'FOR ALL HIDDEN COLUMNS SIZE 1' ); END; /
and re - run your query and post once again the new plan explain him like you did before
Best regards
Mohamed Houri
-
Is it possible to create an index of global partitioned hash on the columns of the primary key of a table that is not partitioned itself?
If a table has a PK constraint there an index generated automatically on the PK columns.
Y at - it a syntax to make the overall index partitioned hash when creating the constraint of PK,.
or syntax to MODIFY the indexes after the creation of the constraint?When you add a Pk to a table, you can choose a place already in the index. Something like
alter table t1 add constraint T_PK primary key (col1) using index T_IDX;
Create the partitioned index as you like with the same columns as the primary key and use the foregoing.
-
ORA-14086: a partitioned index cannot be reconstructed as a whole
Hello
I'm trying to rebuld index had error below.
SQL > alter index rms12. PRICE_HIST_I1 regeneration;
ALTER index rms12. PRICE_HIST_I1 rebuild
*
ERROR on line 1:
ORA-14086: a partitioned index cannot be reconstructed as a whole
Select index_name, separated from dba_indexes where table_name = 'PRICE_HIST '.
o/p
Index_name partitioned
PRICE_HIST_I2 YES
PRICE_HIST_I1 YES
Thank youIf you need to find the names of partition in dba_ind_paritions.
------
Sybrand Bakker
Senior Oracle DBA -
partitioning and partitioned index
I created 1 table with partitions.
CREATE TABLE SAMPLE_ORDERS
(NUMBER OF ORDER_NUMBER,
ORDER_DATE DATE,
NUMBER OF CUST_NUM
NUMBER OF TOTAL_PRICE
NUMBER OF TOTAL_TAX
NUMBER OF TOTAL_SHIPPING)
PARTITION OF RANGE (ORDER_DATE)
(
SO99Q1 PARTITION VALUES LESS THAN (TO_DATE (APRIL 1, 1999 ',' MON-DD-YYYY "")),
SO99Q2 PARTITION VALUES LESS THAN (TO_DATE (1 JULY 1999 ',' MON-DD-YYYY "")),
SO99Q3 PARTITION VALUES LESS THAN (TO_DATE (OCTOBER 1, 1999 ',' MON-DD-YYYY "")),
SO99Q4 PARTITION VALUES LESS THAN (TO_DATE (JANUARY 1, 2000 ',' MON-DD-YYYY "")),
SO00Q1 PARTITION VALUES LESS THAN (TO_DATE (APRIL 1, 2000 ',' MON-DD-YYYY "")),
SO00Q2 PARTITION VALUES LESS THAN (TO_DATE (JULY 1, 2000 ',' MON-DD-YYYY "")),
SO00Q3 PARTITION VALUES LESS THAN (TO_DATE (OCTOBER 1, 2000 ',' MON-DD-YYYY "")),
SO00Q4 PARTITION VALUES LESS THAN (TO_DATE (JANUARY 1, 2001 ',' MON-DD-YYYY ""))
)
;
Few questions is now.
1. How can I create indexes on the table.
2. How can I rebuild the index if it is partitioned index?
3. What is the impact of the reconstruction of indexes on the table (all table locks or just locks partitioned index)
4. If I want to create partition for future purposes, for example for each month do I need to partition manually created for the same thing?
Thank youWhen you create an Index with the keyword LOCAL and do not specify the names of each partition, Oracle uses the default Partition of Table name.
For example, to rebuild the index corresponding to the first partition of the table, partition
alter index sales_orders_ndx_l rebuild partition SO99Q1 ;
DML may continue to run against the other partitions of the table (and their corresponding indices).
When you do the maintenance of the score as the addition of a new Partition or split an existing Partition, for each Partition of the 'new' Table, a corresponding Index Partition will be automatically created.
In order to ensure that the affected Index partitions are not left in a State UNUSABLE but are also rebuilt with maintaining the Table Partition, add the clause INDEX of UPDATE of the ALTER TABLE statement you use. -
Table partition and no partition indexes
Hello
I have the partition table that contains about 1 ml recods and he have daily score.
This partition table have only a unique index that is a partition No.
CREATE UNIQUE INDEX xxxxxx WE yyyyyy
(ITEM_GUID, IMAGE_SIDE)
LOGGING
TABLESPACE zzzzzz
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE)
INITIAL 170M
ACCORDING TO 1 M
MINEXTENTS 1
MAXEXTENTS 2147483645
PCTINCREASE 0
DEFAULT USER_TABLES
)
NOPARALLEL;
I drop very first partition of
Fall of ALTER TABLE wwww.yyyyy SCORE PT_20080706;
But this isn't my unique INVALID index.
Is it normal to have a VALID unique index after the fall of any partition?
Do I have to rebuild a unique index again?
DB: oracle 10.2.0.3
platform: solaris
Thanks in advanceAccording to the Oracle doc, if you drop the bulkhead with an overall index,.
All index global, or all partitions of global partitioned indexes are marked UNUSABLE unless one of the following are true:
You specify the UPDATE INDEX (cannot be specified for tables organized by index. Use GLOBAL updating INDEXES.)
The dropped partition or its subparts are empty
more info here
http://download.Oracle.com/docs/CD/B19306_01/server.102/b14231/partiti.htm#sthref2751Check ALL_PART_INDEXES and ALL_INDEXES to check the State of your indexes.
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Count (*) using partitioned index gives incorrect results
I have a table partitioned by hash with 4 index the.
Table name: store_assortment
clues the: idx1 (master_id), idx2 (store), idx3 (item), idx4 (request_id)
When I run this query result is 13649:
SELECT COUNT (*)
OF store_assortment
WHERE to store = 6010
ORDER BY point;
When I run this query result is 13648:
SELECT COUNT (*)
OF store_assortment
WHERE store = 6010;
I rebuild all indexes, but the results are the same. Can anyone point to a bug or something that can explain this?
I dropped and recreated the indices and values are correct now. Reconstruction did not work.
Thanks for all the help.
-
Hang partitioned index text on construction
I have a partitioned table with 2 000 000 + lines with a blob column. When you try to create an index of type CTXSYS. CONTEXT that is partitioned correctly finished 8 of 10 partitions but crashes right on the other 2 indefinitely. I am building with a parallel degree of 8. Three of these have expected of 'direct path read' while the rest are the "PX Deq: execution Msg.
Any ideas?
Ok. 18121298 patch solves this problem. Discovered this while trying to make a service request.
-
Dear experts,
I do the following:
SQL > create table tab (an integer, whole b)
2 partition () range (a)
3 partition p10 values less than (10),
4 partition p20 values less than (20).
5 partition pmax values less than (10000)
6 );
Table created
SQL > alter table tab add constraint tab_pk key primary (b);
Table changed
SQL > insert into tab values (1, 1);
1 row inserted
SQL > insert into tab values (19: 2);
1 row inserted
SQL > insert into tab values (9999, 3);
1 row inserted
SQL > commit;
Validation complete
SQL > select index_name, status of user_indexes where index-name = "TAB_PK";
INDEX_NAME STATUS
------------------------------ --------
VALID TAB_PK
SQL > tab table split partition of alter
2 pmax (30) into (partition, partition pmax p30);
Table changed
SQL > select index_name, status of user_indexes where index-name = "TAB_PK";
INDEX_NAME STATUS
------------------------------ --------
VALID TAB_PK
Could you explain please why the overall index of tab_pk was not marked UNUSABLE? The partmax of the partition was not empty.
It looks like you were the beneficiary of an optimization: Maintenance of Partitions
-
Split partition - index / local
Hi friends,
I'm trying to divide a table partition.
Please let me know if the syntax below are correct.
If the local index used
ALTER TABLE SNYT. PART_ESTD
ESTD_M13_S22 PARTITION SPLIT TO ('IS', 13, '22')
IN (ESTD_M13_S21 PARTITION, PARTITION ESTD_M13_S22)
update of the index;
(by http://asktom.oracle.com/pls/asktom/f?p=100:11:0:::P11_QUESTION_ID:1401247200346349807)
=================================================================
If used Global indexes
ALTER TABLE SNYT. PART_ESTD
ESTD_M13_S22 PARTITION SPLIT TO ('IS', 13, '22')
IN (ESTD_M13_S21 PARTITION, PARTITION ESTD_M13_S22)
UPDATE GLOBAL INDEXES;
Concerning
KSG
Published by: KSG on 23 August 2012 18:27
Published by: KSG on 23 August 2012 18:51http://docs.Oracle.com/CD/E11882_01/server.112/e26088/statements_3001.htm#i2131218
http://docs.Oracle.com/CD/E11882_01/server.112/e26088/statements_3001.htm#i2151566
-
Table partitioning/indexing strategy
Hello
I have a data warehouse containing medical data. Normally, we develop queries against a small number of patients (e.g. where patient_id < 100) and adjust upward to run on a greater number (usually the entire base of 26 000 patients least some due to exclusion criteria).
The largest table contains about 200 million lines, and I'm working on the question of whether I can change the partitioning and indexing in order to improve the performance of queries.
The table contains 'patient_id', 'visit_id', 'item_id', 'chart_time', 'value' and 'valuenum' column and a variety of other columns that are rarely used. The id id columns are integers, pointrefers to the type of value is saved and value (varchar2) and valuenum (number) contain the data. chart_time is a "timestamp with time zone '.
Currently, there are clues on patient_id, visit_id and item_id and a composite index on (visit_id, itemid). There is no partitioning for the moment.
1. is the current index redundent? that is visit_id, item_id and composite (visit_id, item_id) do I need?
2. we started to run queries of time based for specific patients. that is, we want data for specific items, for a specific set of visits, in the first day. An index on (visit_id, chart_time, item_id) would help with that?
3. the table of patient_id partitioning help? We "never" queries between patient and therefore the data of each patient are independent. The database is running on 2 raid controllers, one for data, one for the index, perhaps it would be better to put half of patients on a single controller, and the other half on the other?
Thank you
Dan ScottHello
1. the index on visit_id sounds redundant as it is the main column in the composite index (visit_id, itemid).
2. added chart_time to the composite index could be useful that the pair of columns (visit_id, item_id) is not quite selective. You must put in place a realistic model of your data distribution, choose some queries that you intend to run often watch their performance. It is not possible to predict performance in function purely thereotical reasoning.
3 bear partitioning functions: maintenance of data (for example truncating, deleting or moving old partitions) and eliminating the irrelevant data in queries (partition size). You imagine you having to truncate, delete, or move a partition that is defined by a range of patient_id? Probably not. Can you imagine the need to restrict your query to a certain range of patient_id? Unlikely. That answers your question.Best regards
Nikolai -
Hi all
Is it possible to partition the partion of maxvalue index created using index range partitioning.Hello
If you talk to split a partition into several smaller, then Yes, you can do, see:
http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:660224234238
If not, please clarify what you mean.
Best regards
Nikolai -
Global partitioned index shows State N/A sqldeveloper
Hi again,
I tried to create a global index divided by 4 hash partitions to be used as the primary key.
I think I understand the syntax for creating an index of global hash (using 4 partitions) that is used by the primary key constraint.
Here's my response:
ALTER table vehicle_data
Add the constraint of KEY PRIMARY C_PK_ID
(
ID
)
USING INDEX
HASH PARTITION (LWVD_ID)
4. THE PARTITIONS
ENABLE;
What I don't understand is this:
When I take the sqldeveloper, go to the table and select the tab 'index' index status is indicated by n/a. When I create the index without partitioning it shows valid.
What is the problem that I forgot?
Thanks again,
AndreasYou forgot that it is partitioned.
If you look in ALL_IND_PARTITIONS for the State.
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Hello
I have created a new local index and found a syntax problem, I thought I had created an index as local, but it was created under the global name.
Here is my syntax of test:
If I run the interviewed next I get results only for the index 'I_IA ':CREATE TABLE invoices (invoice_no NUMBER NOT NULL, invoice_date DATE NOT NULL, comments VARCHAR2(500)) PARTITION BY RANGE (invoice_date) (PARTITION invoices_q1 VALUES LESS THAN (TO_DATE('01/04/2001', 'DD/MM/YYYY')) , PARTITION invoices_q2 VALUES LESS THAN (TO_DATE('01/07/2001', 'DD/MM/YYYY')) , PARTITION invoices_q3 VALUES LESS THAN (TO_DATE('01/09/2001', 'DD/MM/YYYY')) , PARTITION invoices_q4 VALUES LESS THAN (TO_DATE('01/01/2002', 'DD/MM/YYYY'))); CREATE INDEX i_ia ON invoices (comments) local CREATE INDEX i_ia2 ON invoices local (comments)
My production table I thought that I had created a table with local index and only when the fact of the decline of the partitions I realized that something was wrong because it was taking too long to execute.SELECT * FROM DBA_INDEXES, dba_ind_PARTITIONS WHERE TABLE_NAME like UPPER('%invoices%') and DBA_INDEXES.INDEX_NAME = dba_ind_PARTITIONS.INDEX_NAME
My database is:
Oracle Database 11 g Enterprise Edition Release 11.2.0.2.0 - 64 bit Production
Shouldn't receive a syntax warning if "local" is used outside the correct place?
Thank you
Ricardo TomasThis position in the CREATE INDEX Syntax [url http://docs.oracle.com/cd/E11882_01/server.112/e26088/statements_5012.htm] is used for a table alias.
LOCAL is not a reserved word.
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