Global partitioned index hash

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.

Tags: Database

Similar Questions

  • 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,
    Andreas

    You forgot that it is partitioned.

    If you look in ALL_IND_PARTITIONS for the State.

  • 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 advance

    According 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#sthref2751

    Check ALL_PART_INDEXES and ALL_INDEXES to check the State of your indexes.

  • with hash index global partition

    Hello

    I use SQL DEVELOPER DATA MODELER Version 4.0.0.833.

    In the database have an index with global hash partitions (11 g Enterprise Edition Release 11.2.0.2.0)

    When I sincronize with the database and the search for differences I find this tablespace in metadata comes from DDBB is null.

    Thank you for all.

    David

    Hi David,

    It seems that the storage space for the partitions of a partitioned index global hash is not currently imported from the database.

    I connected an enhancement request to support this feature.

    David

  • Partitioned index not used in the query

    Hello

    Oracle 10.2.0.1
    Windows xp

    I'm confused why oracle did not use a partitioned below index:
    SQL> set line 200;
SQL> SET AUTOTRACE TRACEONLY EXPLAIN;
SQL> select * from MYTABLE where PROD like 'SOAP%';

Execution Plan
----------------------------------------------------------
Plan hash value: 2899783245

-------------------------------------------------------------------------------------------------------
| Id  | Operation              | Name         | Rows  | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
-------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT       |              |  1852 |   115K|   992   (3)| 00:00:12 |       |       |
|   1 |  PARTITION RANGE SINGLE|              |  1852 |   115K|   992   (3)| 00:00:12 |     8 |     8 |
|*  2 |   TABLE ACCESS FULL    |      MYTABLE |  1852 |   115K|   992   (3)| 00:00:12 |     8 |     8 |
-------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   2 - filter("PROD" LIKE 'SOAP%')

SQL> desc MYTABLE;
 Name                                                                                  Null?    Type
 ------------------------------------------------------------------------------------- -------- -----------
 PROD_ID                                                                               NOT NULL VARCHAR2(7)
 PROD                                                                                  NOT NULL VARCHAR2(30)
 AREA                                                                                  NOT NULL VARCHAR2(30)
 SUB_AREA                                                                              NOT NULL VARCHAR2(30)

SQL>
    But when I said something different; then it accesses the indexes:
    SQL> select count(*) from MYTABLE where PROD like 'SOAP%';

Execution Plan
----------------------------------------------------------
Plan hash value: 1393798969

----------------------------------------------------------------------------------------------------
| Id  | Operation               | Name     | Rows  | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
----------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT        |          |     1 |    16 |    18   (0)| 00:00:01 |       |       |
|   1 |  SORT AGGREGATE         |          |     1 |    16 |            |          |       |       |
|   2 |   PARTITION RANGE SINGLE|          |  1852 | 29632 |    18   (0)| 00:00:01 |     8 |     8 |
|*  3 |    INDEX RANGE SCAN     | PROD_IDX |  1852 | 29632 |    18   (0)| 00:00:01 |     8 |     8 |
---------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   3 - access("PROD" LIKE 'SOAP%')
       filter("PROD" LIKE 'SOAP%')

SQL>
    How to create prod_idx:
    SQL> ED
Wrote file afiedt.buf

  1  create index prod_idx ON mytable(prod)
  2  global partition by range(prod)
  3  (partition prod1 values less than ('A'),
  4  partition prod2 values less than ('B'),
  5  partition prod3 values less than ('C'),
  6  partition prod4 values less than ('D'),
  7  partition prod5 values less than ('E'),
  8  partition prod6 values less than ('F'),
  9  partition prod7 values less than ('G'),
 10  partition prod8 values less than ('H'),
 11  partition prod9 values less than ('I'),
 12  partition prod10 values less than ('J'),
 13  partition prod11 values less than ('K'),
 14  partition prod12 values less than ('L'),
 15  partition prod13 values less than ('M'),
 16  partition prod14 values less than ('N'),
 17  partition prod15 values less than ('O'),
 18  partition prod16 values less than ('P'),
 19  partition prod17 values less than ('Q'),
 20  partition prod18 values less than ('R'),
 21  partition prod19 values less than ('S'),
 22  partition prod20 values less than ('T'),
 23  partition prod21 values less than ('U'),
 24  partition prod22 values less than ('V'),
 25  partition prod23 values less than ('W'),
 26  partition prod24 values less than ('X'),
 27  partition prod25 values less than ('Y'),
 28  partition prod26 values less than ('Z'),
 29* partition prod27 values less than (MAXVALUE))
SQL> /

Index created.
    SQL> select count(*) from MYTABLE;

Execution Plan
----------------------------------------------------------
Plan hash value: 3323402158

--------------------------------------------------------------------------------------------------------------
| Id  | Operation                      | Name                | Rows  | Cost (%CPU)| Time     | Pstart| Pstop |
--------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT               |                     |     1 |   311   (0)| 00:00:04 |       |       |
|   1 |  SORT AGGREGATE                |                     |     1 |            |          |       |       |
|   2 |   PARTITION RANGE ALL          |                     |    14M|   311   (0)| 00:00:04 |     1 |    27 |
|   3 |    BITMAP CONVERSION COUNT     |                     |    14M|   311   (0)| 00:00:04 |       |       |
|   4 |     BITMAP INDEX FAST FULL SCAN| MS_IDX_MYTABLE      |       |            |          |     1 |    27 |
--------------------------------------------------------------------------------------------------------------
    Table is to have 14693792 lines.

    Why this is the case, please guide me to understand. Thank you.

    Published by: user12050217 on July 29, 2010 23:37

    Published by: user12050217 on July 29, 2010 23:38

  • 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.

  • Index hash

    Hello

    What is index hash, to which cases to use hash indexes in oracle. And please give me a simple example.

    Thanks in advance,
    Sanjeev.

    >
    What is index hash, to which cases to use hash indexes in oracle. And please give me a simple example.
    >
    You'll never HAVE to use a hash, or any, of the index. You only use an index if it offers some advantages.

    A hash index is often used when there is a lot of data that you want to break into subsections or separate partitions or when you have a set of distinct values that you want to partition on and will not have to specify all partitions individually.

    An example would be if you have data for the 50 States of United States and you want to partition by State. You can create a hash table partitioned on STATE_NAME or STATE_CODE (50 distinct values) and you would get probably 50 scores. It would be easier than creating a partitioned table list and have to specify different 50 "lists" (one for each State).

    Oracle would still be partition pruning if your filter predicates used queries such as: WHERE STATE_NAME = "CALIFORNIA". A single partition would be used in the query, even if you use hash partitioning. It is because the word 'CALIFORNIA' always in the same value and Oracle can make this hash when the query is compiled and determine that one partition is required.

  • PARTITION OF HASH ON LIST

    version: Oracle 10g

    Hi all
          i have a table partitioned by hash(8 segments) with 200 millions records . now we want to move to list partition the same table by state.
      Is there a way to drop hash Partitions with out losing the data, and Add list partition.
     
    

    No, you can't 'convert' hash Partitions to list of Partitions.

    You must create an empty table that is partitioned from the list and then hover over the data to the new table.
    You can use Direct path Insert (i.e. INSERT with APPEND tip) with degree PARALLEL and NOLOGGING on the target table to speed up the copy.
    You can also use DBMS_REDEFINITION to migrate the data.
    In both cases, you must also create the index on the new table.

    Hemant K Collette

  • 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

  • Partitioned index

    Using Oracle 11.2.0.3

    We 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 11.2.0.3

    We 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

  • Local partitioned indexes

    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 21:51

    Dear 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

  • 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 you

    If 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 you

    When 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.

  • DML on global partition

    Hi gurus,

    When truncate us partition, including the overall index, there is a clause 'updated global index", which updates the index.

    For the local index, as much as I understoond, oracle automatically maintains according to the partitioned table.

    Now for DML have we any method that would have done the same thing?

    As far as I understand for DML on the indexed column, index gets each time reconstruction.

    Isn't that true for all indexes. Could someone explain to me how this reconstruction happens and how long would it take?

    Does it affect your query while the index is in the process of reconstruction?

    Thank you

    When truncate us partition, including the overall index, there is a clause 'updated global index", which updates the index.

    No - the clause is INDEX of UPDATE. You only use 'update global indexes' for a table held in index.

    Partition management

    For the local index, as much as I understoond, oracle automatically maintains according to the partitioned table.

    Now for DML have we any method that would have done the same thing?

    As far as I understand for DML on the indexed column, index gets each time reconstruction.

    Non - index do NOT get every time rebuilt. They are 'held' by Oracle. Perform the DML on a line, update the index of this line.

  • 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 18:51

    http://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

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