Global index unpartitioned reconstruction shot

Similar Questions

• Select Insert on the same partition: RELATIONAL() - REMOVE a GLOBAL INDEX

  Hi all

  I found on v$ sqlarea suite statament:

  insert / * + / RELATIONAL PARALLEL ("TABLENAME") ("TABLENAME", 1) ADD NESTED_TABLE_SET_SETID NO_REF_CASCADE * /

  in "SCHEMA." "" TABLENAME ' partition ('DAY20151015')

  (

  Select / * + RELATIONAL("TABLENAME") PARALLEL("TABLENAME", 1) * /.

  *

  a 'schema '. " TABLENAME' partition ("DAY20151015") ".

  )

  Remove the global index

  This is a query that reads and writes the same data on the same partition!

  I think that it is generated by Oracle, but I don't undestand what he does.

  Can you give me some explanation or assistance on this query?

  Thank you very much

  This is a query that reads and writes the same data on the same partition!

  Well not - enough is NOT that.

  The RELATIONAL indicator causes unnest Oracle object data and insert the data from the attribute itself rather than the object.

  The indicators used are usable only by Oracle-c ' is why they are undocumented. You can find songs on the web, but practically everything you find is "best estimate".

  Oracle SQL tips - other indicators [HelloDBA.COM]

• Local and Global Index confusion...

  Hi all

  Oracle Database 10 g Enterprise Edition Release 10.2.0.3.0 - 64 bit

  I have a partitioned table, when developing the application, they have committed an index ON this table. Previously, whenever we use to run 'Alter Table add partition..' global index becomes unusable and I used to rebuild the same. But in recent months, each time we add partition (Alter Table... ADD Partition..) on this table, index will get unusable to stat instead it shows that indexes THE data dictionaries (Lacality = Local in USER_PART_INDEXES). Partitioned indexes are also automatically created with each table partition creation.

  But the query below shows the output indexes are GLOBAL.

  Select double dbms_metadata.get_ddl('INDEX','ABCD','USER_A');

  Can someone me help on what actually happened here.

  But the example above has the keyword LOCAL?

  CREATING INDEX 'SEB '. "" CDRV_I1 ""SEBS ". "" CDRV ' ('CDRV_CURNT_FL', 'IND_LAST_VERSN', 'GMT_SEIZ_DT_TIME', 'CLECT_ZONE', 'CALL_ID',

  "SWTCH_ADMIN_ABBR")

  PCTFREE, INITRANS 10 2 MAXTRANS 255 LOGGING

  STORAGE (INITIAL 10485760 NEXT 10485760 MINEXTENTS 2 MAXEXTENTS 2147483645)

  PCTINCREASE 0 DEFAULT USER_TABLES)

  "SEBS_CDRV_TAB" of LOCAL TABLESPACE

  (PARTITION "CDRV_HISTORIC"

  PCTFREE, INITRANS 10 2 MAXTRANS 255

  STORAGE (INITIAL 10485760 NEXT 10485760 MINEXTENTS 2 MAXEXTENTS 2147483645)

  PCTINCREASE 0 FREELISTS 1 FREELIST GROUPS 1 DEFAULT USER_TABLES)

  TABLESPACE "SEBS_CDRV_TAB."

  If it looks like a duck and quacks like a duck, it is very probably a duck (or the local index)

  See you soon

  Richard Foote

  http://richardfoote.WordPress.com/

• conversion index global index local

  Hello guys, someone tell me how to convert a global index of a local index.
  The only method I know is to
  1 remove the global index
  2. create the index even with the local keyword.
  
  I was wondering is there any alter index statement which allows you to change the overall index to a local index?
  
  Thank you

  969224 wrote:
  Hello guys, someone tell me how to convert a global index of a local index.
  The only method I know is to
  1 remove the global index
  2. create the index even with the local keyword.

  I was wondering is there any alter index statement which allows you to change the overall index to a local index?

  Thank you

  There is no such statement

• Local/Global index for OLTP

  In our application (OLTP), we have a rule on
  
  * "" on all the partitioned tables, we need to have just partitioned LOCAL index "... *"
  I think that this rule is valid, because when we had GLOBAL index partitioned on the table when concurrent transactions
  was going on, we got below error:
  ORA - 00054:resource busy and...
  We get this error when rebuild us all indexes not valid...
  
  When we changed all the GLOBAL partitioned index at the LOCAL level... we get this error...
  
  We use MEV for cloud computing (i.e. we have several tenants and a partition for each tenant)
  
  
  My doubt is, is the rule as mentioned above may be still valid, because it is at odds with what is mentioned in the Oracle
  Documentation (the concepts guide)...
  http://docs.Oracle.com/CD/B19306_01/server.102/b14220/partconc.htm#i461446
  
  '' In general, you should use an index for OLTP applications and local storage of data or applications of DSS ''
  
  We use oracle 10 g 2...

  In our case, research www.lesormes.com will not create several partitions, only one partition (we have a partition for each tenant).

  Local is even better for you.

  Thanks for suggesting to rebuild online (for global)... I'll try with that... .but he would have any problem of performance relative to the index LOCAL...

  I don't think abnd so there shouldn't be any performance impact, rather it improves performance in locaking not all the table exclusively for the index creaiton.

  NY suggestion/idea why Oracle recommends partitioned Global index for OLTP?

  Because normally we have much aprtition of drop/merger/split partition of things in OLTP type (these operations are important in DSS normally loading data).

  any suggestion/idea why Oracle recommends partitioned Global index for OLTP?

  Because they are the fastest to access a record of a table and OLTP, response time is what counts. If you compare your reports OLTP or transactions, to spawn multiple partitions (a small percentage could access only a single partition and rules are made for the majority of cases, not for the small number of cases) and global index are better in this scenario.
  at MAS, you always have a huge data and as partition operation may stop your work until the index is rebuilt (which can take several hours for the huge data) where oracle is recommended to have a local indexing policy.

  Salman

• Check the local or global index

  Hi all
  I wanted to know how we confirm/check if an index is local or global on a system of oracle 10 g 2
  
  Thank you
  Rossy.Rocs

  Lacotte,

  Small correction was need here that the partitioned dba_indexes column will only be "indicates whether the index is partitioned (YES) or not (NO); While op asks how do I know if it is a local or global index. So for that he must check the below command:

  SELECT INDEX_NAME, THE LOCALITY OF USER_PART_INDEXES;

  Column of the locality of user_part_indexes ' indicates if the partitioned index is local (LOCAL) or global (GLOBAL).
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14237/statviews_2015.htm#REFRN20156

  Concerning
  Girish Sharma

• Partitioned global index on partitioned table range, but the index partition does not work

  Hello:

  I was creating an index partitioned on table partitioned and partitioned index does not work.

  create table table_range)

  CUST_FIRST_NAME VARCHAR2 (20).

  CUST_GENDER CHAR (1),

  CUST_CITY VARCHAR2 (30),

  COUNTRY_ISO_CODE CHAR (2),

  COUNTRY_NAME VARCHAR2 (40),

  COUNTRY_SUBREGION VARCHAR2 (30),

  PROD_ID NUMBER NOT NULL,

  CUST_ID NUMBER NOT NULL,

  TIME_ID DATE NOT NULL,

  CHANNEL_ID NUMBER NOT NULL,

  PROMO_ID NUMBER OF NON-NULL,

  QUANTITY_SOLD NUMBER (10.2) NOT NULL,

  AMOUNT_SOLD NUMBER (10.2) NOT NULL

  )

  partition by (range (time_id)

  lower partition p1 values (u01 tablespace to_date('2001/01/01','YYYY/MM/DD')),

  lower partition (to_date('2002/01/01','YYYY/MM/DD')) tablespace u02 p2 values

  );

  create index ind_table_range on table2 (prod_id)

  () global partition range (prod_id)

  values less than (100) partition p1,

  lower partition p2 values (maxvalue)

  );

  SQL > select TABLE_NAME, SUBPARTITION_COUNT, HIGH_VALUE, nom_partition NUM_ROWS of user_tab_partitions;

  TABLE_NAME NOM_PARTITION SUBPARTITION_COUNT HIGH_VALUE NUM_ROWS

  ----------- ---------------- ------------------ -------------------------------------------------------------------------------- ----------

  TABLE_RANGE P2 0 TO_DATE (' 2002-01-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIA 259418)

  TABLE_RANGE P1 0 TO_DATE (' 2001-01-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIA 659425)

  SQL > select INDEX_NAME, NUM_ROWS nom_partition, HIGH_VALUE user_ind_partitions;

  INDEX_NAME NOM_PARTITION HIGH_VALUE NUM_ROWS

  ------------------------------ ------------------------------ -------------------------- ----------

  P1 IND_TABLE_RANGE 100 479520

  IND_TABLE_RANGE P2 MAXVALUE 439323

  SQL > EXECUTE DBMS_STATS. GATHER_TABLE_STATS (USER, 'TABLE_RANGE');

  SQL > EXECUTE DBMS_STATS. GATHER_TABLE_STATS (USER, 'TABLE_RANGE', GRANULARITY = > 'PARTITION');

  SQL > EXECUTE DBMS_STATS. GATHER_INDEX_STATS (USER, 'IND_TABLE_RANGE');

  SQL > EXECUTE DBMS_STATS. GATHER_INDEX_STATS (USER, 'IND_TABLE_RANGE', GRANULARITY = > 'PARTITION');

  SQL > set autotrace traceonly

  SQL > alter shared_pool RAS system;

  SQL > changes the system built-in buffer_cache;

  SQL > select * from table_range

  where prod_id = 127;

  ---------------------------------------------------------------------------------------------------

  | ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

  ---------------------------------------------------------------------------------------------------

  |   0 | SELECT STATEMENT |             | 16469 |  1334K |  3579 (1) | 00:00:43 |       |       |

  |   1.  RANGE OF PARTITION ALL THE |             | 16469 |  1334K |  3579 (1) | 00:00:43 |     1.     2.

  |*  2 |   TABLE ACCESS FULL | TABLE_RANGE | 16469 |  1334K |  3579 (1) | 00:00:43 |     1.     2.

  ---------------------------------------------------------------------------------------------------

  Information of predicates (identified by the operation identity card):

  ---------------------------------------------------

  2 - filter ("PROD_ID" = 127)

  Statistics

  ----------------------------------------------------------

  320 recursive calls

  2 db block Gets

  13352 consistent gets

  11820 physical reads

  0 redo size

  855198 bytes sent via SQL * Net to client

  12135 bytes received via SQL * Net from client

  1067 SQL * Net back and forth to and from the client

  61 sorts (memory)

  0 sorts (disk)

  15984 rows processed

  Once the sentence you say ' does not ' and then you go to paste plans that seem to show that it "works".

  What gives?

  In fact, if you look at the plans - think Oracle you have 16 k rows in the table and he'll be back k 12 rows for your select statement. In this case, Oracle is picking up the right plan - full scan 16 ranks of k is a lot less work to digitize the index lines k 12 followed by the research of rank k 12 rowid.

• How do I know it's global index or local index?

  Hi all
  
  How will I know if it's overall index or local index?
  
  Thank you!
  
  Dan.

  Salvation;

  Check defination of index? Overall index below has defination

  CREATE INDEX month_ix ON sales (sales_month)
  GLOBAL PARTITION BY RANGE (sales_month)

  For more information, please visit:

  http://download.Oracle.com/docs/CD/B13789_01/server.101/b10739/partiti.htm

  Respect of
  HELIOS

• Local partitioned Global Index Index conversion online.

  Hi all
  I use the Release of oracle 10.2.0.4.0 version. As a solution to solve a performance problem, I intend to convert one of the partitioned local index in overall index. Here is the script that I made, but it will take a time out as because other applications may use the same index and will suffer from performance during the transition problem. My question is if I can reach the same onlline, only because the prod database is too busy down time? Please suggest.
  
  Script:
  
  drop index INDX_c1c2;
  
  create the IDX_c1c2 index on tab1 (c1, c2);

  930254 wrote:
  Hi all
  I use the Release of oracle 10.2.0.4.0 version. As a solution to solve a performance problem, I intend to convert one of the partitioned local index in overall index. Here is the script that I made, but it will take a time out as because other applications may use the same index and will suffer from performance during the transition problem. My question is if I can reach the same onlline, only because the prod database is too busy down time? Please suggest.

  Script:

  drop index INDX_c1c2;

  create the IDX_c1c2 index on tab1 (c1, c2);

  Your application will run without index during the time it takes to create. An alternative approach which avoids this would be:

  create index INDX_c1c2null on tab1(c1,c2,null);
  drop index INDX_c1c2;
  

  --
  John Watson
  Oracle Certified Master s/n
  http://skillbuilders.com

  Published by: JohnWatson on February 17, 2013 11:35
  Typo - forgot to change the name of the index

• partation local and global index

  Hello
  
  I want partation tables in my database.
  Can someone tell me what exactly the difference between the local index & glabal.
  is there any link which can give me more informetion about the same.
  
  Thank you

  Local index is the index that is created on the partition key column name.

  Hi Anand

  Just to be clear, that a local index must not necessarily be created on partitioned key column, it is simply automatically partitioned in the same way as the underscore.

  If it contains effectively partitioned key columns is completely optional.

  See you soon

  Richard Foote
  http://richardfoote.WordPress.com/

• Global and local index with respect to table partitioned

  Hello
  I am very confused about local and global index in a partitioned table.
  In addition, local index can be partitioned and unpartitioned.
  In addition, what is the difference between global partitioned and unpartitioned
  the index?
  
  As far as I know,
  If there is local index then there will be as many partitions as the partition table
  and local parition will index contains the index of only the lines of their corresponding scores.
  In addition, the local index maintananence is easier because drop us or create a partition, only index corresponding partition is affected then that in the case of the overall index,
  If we create a new partition of the table, to rebuild the global index.
  Am I wrong?
  
  Concerning

  Hello

  This means that when you add/drop/split/merge table any partition (maintaiance) lying below, its corresponding partition in index is automatically added/deleted/split/merged. Oracle takes care of it internally without user intervention.

  Concerning
  Anurag

• 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.

• index global vs local

  Hello world
  What is the diff b/w local and global indexes, in what situation we wil use these two?

  CREATING ORACLE INDEXES. Laughing out loud. OMG_1
  (DEFINITION_ID)
  TABLESPACE WTF
  INITRANS 2
  MAXTRANS 255
  LOGGING
  LOCAL)
  P20100604 PARTITION
  LOGGING
  NOCOMPRESS
  TABLESPACE WTF
  PCTFREE 10
  INITRANS 2
  MAXTRANS 255
  STORAGE)
  ORIGINAL 128 K
  NEXT 128K
  MINEXTENTS 1
  MAXEXTENTS 2147483645
  PCTINCREASE 0
  DEFAULT USER_TABLES
  ),
  P20100605 PARTITION
  LOGGING
  NOCOMPRESS
  TABLESPACE WTF
  PCTFREE 10
  INITRANS 2
  MAXTRANS 255
  STORAGE)
  ORIGINAL 128 K
  NEXT 128K
  MINEXTENTS 1
  MAXEXTENTS 2147483645
  PCTINCREASE 0
  DEFAULT USER_TABLES
  )
  )
  NOPARALLEL;

  CREATE A UNIQUE ORACLE INDEX. Laughing out loud. OMG_2
  (FOLDER_ID)
  LOGGING
  TABLESPACE WTF
  PCTFREE 10
  INITRANS 2
  MAXTRANS 255
  STORAGE)
  80K INITIAL
  NEXT 128K
  MINEXTENTS 1
  MAXEXTENTS 2147483645
  PCTINCREASE 0
  DEFAULT USER_TABLES
  )
  NOPARALLEL;

• status indication by n/a, after the reconstruction for all subparts

  Hello Experts,

  I created a table and apply composite partition (partion and sub list partition range). Created local indexes on multiple columns and rebuilt indexes on subpartitions successfully.

  After reconstruction in dba_ind_subpartitions I see only index status changed to USABLE.

  (1) after reconstruction of the index why the State always shows by n/a dba_ind_partitions and dba_indexes?

  Please help me if I missed all the steps during the reconstruction?

  See you soon,.

  -Suri

  It is expected, Local index for a partitioned table will indicate the status of dba_ind_partitions (normal partition) or dba_ind_subpartitions(composite partition). N/A does not mean unusable, you arrive seen appropriate.

  DBA_INDEXES will show the status of the index standard and if it has not partitioned global index.

• 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.