Global partitioned index shows State N/A sqldeveloper

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• 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.

• 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
  

• 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.

• 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

• 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

• create partitioned indexes

  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:

  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)

  If I run the interviewed next I get results only for the index 'I_IA ':

  SELECT * 
  FROM DBA_INDEXES,
       dba_ind_PARTITIONS
  WHERE TABLE_NAME like UPPER('%invoices%') and
        DBA_INDEXES.INDEX_NAME = dba_ind_PARTITIONS.INDEX_NAME

  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.
  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 Tomas

  This 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.

• Rebuild partitioned indexes

  Hello
  I have a newbie question.
  I got a clue who have 36 partitions and subpartitions 12996.
  How I can rebuid this index?
  Thank you

  Why you rebuild?

  You can generate the instructions of reconstruction executes a something similar to the following select statement - NOTE that I DID NOT WRITE this AND haven't TESTED it

  SELECT 'ALTER INDEX | index_name | 'REBUILD ONLINE;'
  From user_indexes
  WHERE index-name NOT IN (SELECT index_name FROM user_ind_partitions)
  AND nom_tablespace = 'BI_MIFID_STG_IND. '
  UNION
  SELECT "ALTER INDEX".
  || index_name
  || "REBUILD THE PARTITION.
  || nom_partition
  || 'ONLINE';
  Of user_ind_partitions
  WHERE index-name NOT IN (SELECT index_name FROM user_ind_subpartitions)
  AND nom_tablespace = 'BI_MIFID_STG_IND. '
  UNION
  SELECT "ALTER INDEX".
  || index_name
  || "REBUILD SUBPARTITION.
  || subpartition_name
  || 'ONLINE';
  Of user_ind_subpartitions
  WHERE subpartition_name IN (SELECT nom_partition
  FROM WHERE user_segments
  WHERE nom_tablespace = 'BI_MIFID_STG_IND')
  AND nom_tablespace = 'BI_MIFID_STG_IND. '

• 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.