TIMESTAMP (6) partitioned key-> range partitioned table ddl needed

Similar Questions

• Unique key on the partitioned table range

  Hello
  
  We use a range of composite range-hash partitioned table
  
  Use index - try to do this have same tablespace to the local partitions but not liking it
  
  ALTER table RETAILER_TRANSACTION_COMP_POR
  Add primary key constraint RETAILER_TRANSACTION_COMP_PK (DWH_NUM)
  using index
  LOCAL
  
  
  ORA-14039: partitioning columns must be a subset of the columns in a unique index key
  
  Without local then fine but does not have same tablespace to walls and don't want to make this part of the partition key.
  
  Range Tbal partitioned - it's just a UK to avoid duplicates

  [oracle@localhost ~]$ oerr ora 14039
  14039, 00000, "partitioning columns must form a subset of key columns of a UNIQUE index"
  // *Cause:  User attempted to create a UNIQUE partitioned index whose
  //          partitioning columns do not form a subset of its key columns
  //          which is illegal
  // *Action: If the user, indeed, desired to create an index whose
  //          partitioning columns do not form a subset of its key columns,
  //          it must be created as non-UNIQUE; otherwise, correct the
  //          list of key and/or partitioning columns to ensure that the index'
  //          partitioning columns form a subset of its key columns
  

• NULL values in partition Table of Partition key

  Hi all

  I'm using Oracle 11.2.0.3.

  I can add a null value on partition of Partition table key.

  If so, how?

  Kind regards.

  • You cannot specify NULL in the VALUES clause.

  Referring to clauses PARTITION you set - not the data that you insert.

  If you cannot define a range partition that uses the VALUES LESS THAN and use null to this division. But you CAN use NULL values in the VALUES of integration clause.

  DROP TABLE mytable;

  CREATE TABLE MyTable

  (

  CREATION_DATE DATE,

  Identification number

  )

  PARTITION BY RANGE (CREATION_DATE)

  (

  PARTITION prior_to_2016 VALUES LESS THAN (TO_DATE ('2016-01-01', 'YYYY-MM-DD'));

  LESS THAN PARTITION of VALUES undesirable (MAXVALUE)

  )

  Insert into myTable values (null, null)

  You can use a VALUES clause to insert NULL values - this line will go into the MAXVALUE partition.

• Local primary key on reference partitioned Table

  Oracle running on Red Hat Linux Rel6 11.2.0.3.

  I'm on a closed network, so the following must be typed manually.

  I have a table of documents which is essentially the following:

  (entire annual <-primary key)

  whole Source_ID,

  load_dt date,

  date of doc_dt,

  doc_info clob,

  ...)

  which is partitioned on column source_id.

  I created a partitioned table of reference as follows:

  create table doc_entities_prt)

  whole doc_entity_id

  all annual,

  whole entity_id,

  forced doc_ent_fk (annual) references to documents (annual)

  tablespace...

  allow the movement of the line

  benchmark score (doc_ent_fk);

  The annual column in the child table is not unique.  The doc_entity_id column is unique.  I want to set the primary key on doc_entities_prt as a local index on the doc_entity_id column.  I was not able to find the proper syntax to get there, and now I'm wondering if this is even possible? Any ideas appreciated.

  Oops, I missed part that you want to base the PK of this index. No, it is not possible and has nothing to do with the partitoning reference. A unique index can be partitioned only if it includes a partitioning column. You can create a non-unique without partitioning column partitioned index, but then you can't create PK supported by such an index for the same reason - oracle would not be able to verify uniquenes based on this index partition and controls of the cross-partition are not supported:

  SQL > create table documents)
  2 whole annual,
  3 whole source_id,
  load_dt date 4.
  date of doc_dt 5.
  6 doc_info clob
  7                        )
  8 partition by range (source_id)
  9      (
  10 partition p1 values less than (10),
  11 partition p2 values less than (100)
  12)
  13.

  Table created.

  SQL > create index unique documents_pk
  2 on documents (annual)
  (3) local
  partition 4 p1,
  5 partition p2
  6         )
  7.
  on documents (annual)
  *
  ERROR on line 2:
  ORA-14039: partitioning columns must be a subset of the columns of a unique key
  index

  SQL > create index documents_pk
  2 on documents (annual)
  (3) local
  partition 4 p1,
  5 partition p2
  6         )
  7.

  The index is created.

  SQL > alter table documents
  2 Add the constraint documents_pk
  3 key (annual) elementary school
  4 using index documents_pk
  5.
  change the documents table
  *
  ERROR on line 1:
  ORA-14196: specified Index cannot be used to apply the constraint.

  SQL >

  SY.

• Size of partition on interval range partitioned Table happens when SYSDATE is used in a Where Clause

  We have tables interval range partitioned on a DATE, with a partition for each day column - very standard and straight out of doc Oracle.

  A 3rd party application queries the tables to find the number of rows based on the date range that is located on the column used for the partition key.

  This application uses the date range specified from the current date - i.e. for last two days would be «...» StartDate > SYSDATE-2 "- but the partition size is irrelevant and the explain command plan shows that each partition is included."

  In presenting the request uses the date in a variable partition size location and query table is obviously much better.

  DB is 11.2.0.3 on RHEL6, and default settings - i.e. nothing that could influence the behavior of the optimizer to something unusual.

  I can't work on why this would be the case. It is very easy to reproduce with cases of simple test below.

  I would be very interested to hear any views on why it's that way and if anything can be done to allow the size of the partition to work with a query including SYSDATE because it would be difficult to get the application code has changed.

  In addition to make a case to change the code, I need an explanation of why query using SYSDATE is not advisable and I know this information.

  (1) create a simple partitioned table

  CREATETABLE part_test
     (id                      NUMBER NOT NULL,
      starttime               DATE NOT NULL,
      CONSTRAINT pk_part_test PRIMARY KEY (id)) 
  PARTITION BY RANGE (starttime) INTERVAL (NUMTODSINTERVAL(1,'day')) (PARTITION p0 VALUES LESS THAN (TO_DATE('01-01-2013','DD-MM-YYYY')));
  

  (2) fill in the rows of the table 1 million distributed among 10 partitions

  BEGIN
      FOR i IN 1..1000000
      LOOP
          INSERT INTO part_test (id, starttime) VALUES (i, SYSDATE - DBMS_RANDOM.value(low => 1, high => 10));
      END LOOP;
  END;
  /
  EXEC dbms_stats.gather_table_stats('SUPER_CONF','PART_TEST');
  

  (3) to query the Table of data from the last 2 days using SYSDATE in paragraph

  EXPLAIN PLAN FOR 
  SELECT  count(*) 
  FROM    part_test
  WHERE   starttime >= SYSDATE - 2;
  

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

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

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

  |   0 | SELECT STATEMENT |           |     1.     3 ×  7895 (1) | 00:00:01 |       |       |

  |   1.  GLOBAL TRI |           |     1.     3 ×            |          |       |       |

  |   2.   RANGE OF PARTITION ITERATOR.           |   111K |   867K |  7895 (1) | 00:00:01 |   KEY | 1048575.

  |*  3 |    TABLE ACCESS FULL | PART_TEST |   111K |   867K |  7895 (1) | 00:00:01 |   KEY | 1048575.

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

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

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

  3 - filter("STARTTIME">=SYSDATE@!-2)

  (4) now do the same query, but with SYSDATE - 2 presented as a literal value.

  This query returns the same response but very different cost.

  EXPLAIN PLAN FOR
  SELECT count(*) 
  FROM part_test
  WHERE starttime >= (to_date('23122013:0950','DDMMYYYY:HH24MI'))-2;
  

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

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

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

  |   0 | SELECT STATEMENT |           |     1.     8.   131 (0) | 00:00:01 |       |       |

  |   1.  GLOBAL TRI |           |     1.     8.            |          |       |       |

  |   2.   RANGE OF PARTITION ITERATOR.           |   111K |   867K |   131 (0) | 00:00:01 |   356 . 1048575.

  |*  3 |    TABLE ACCESS FULL | PART_TEST |   111K |   867K |   131 (0) | 00:00:01 |   356 | 1048575.

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

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

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

  3 filter ("STARTTIME" > = TO_DATE (' 2013-12-21 09:50 ',' syyyy-mm-dd hh24:mi:ss'))))

  Thank you in anticipation

  Jim

  sysdate is not constant and its value may change from time optimization of execution; but the optimizer can deduce that he will be a known running value and then produce a plan that will make the size of partition running.  This is the meaning of 'KEY' in the column pstart – the partition of departure will be known at run time and the size of partition will take place in order to eliminate the previous partitions.

  Concerning

  Jonathan Lewis

• Try to convert the partitioned Table of interval in the range... Swap partition...

  Requirement:
  
  Interval of replacement partitioned Table by range partitioned Table
  

  DROP TABLE A;
  
  CREATE TABLE A
  (
     a              NUMBER,
     CreationDate   DATE
  )
  PARTITION BY RANGE (CreationDate)
     INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
     (PARTITION P_FIRST
         VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'));
  
  
  INSERT INTO A
       VALUES (1, SYSDATE);
  
  INSERT INTO A
       VALUES (1, SYSDATE - 30);
  
  INSERT INTO A
       VALUES (1, SYSDATE - 60);

  I need to change this partitioned Table apart to a partitioned range Table. I can do using the EXCHANGE PARTITION. Like if I use the classic method to create another table range partitioned, then:
  
  

  DROP TABLE A_Range
  CREATE TABLE A_Range
  (
  a NUMBER,
  CreationDate DATE
  )
  PARTITION BY RANGE (CreationDate)
     (partition MAX values less than (MAXVALUE));
  
  Insert  /*+ append */  into A_Range Select * from A; --This Step takes very very long..Trying to cut it short using Exchange Partition.

  Problems:
  
  I can't do
  

   ALTER TABLE A_Range
    EXCHANGE PARTITION MAX
    WITH TABLE A
    WITHOUT VALIDATION;
   

  ORA-14095: ALTER TABLE CHANGE requires a not partitioned table nonclustered
  This is because the tables are partitioned. So it does not allow me.
  
  If I instead:
  
  
  Create a table that is not partitioned for exchanging data by partition.
  

    Create Table A_Temp as Select * from A;
    
     ALTER TABLE A_Range
    EXCHANGE PARTITION MAX
    WITH TABLE A_TEMP
    WITHOUT VALIDATION;
     
    select count(*) from A_Range partition(MAX);
   

  -The problem is that all the data is in MAX Partition.
  Even after the creation of a large number of partitions by walls of separation, the data is still in MAX Partition only.
  
  So:
  
  -What we cannot replace a partitioned Table to the Table partitioned using the EXCHANGE PARTITION range interval. that is, we have to insert in...
  -We can do it, but I'm missing something here.
  -If all the data is in MAX Partition due to "WITHOUT VALIDATION", can say us be redistributed in the right type of range partitions.

  You must pre-create the partitions in a_range and then swap one for one for a tmp, and then to arange. With the help of your sample (thanks to proviing code, incidentally).

  SQL> CREATE TABLE A
    2  (
    3     a              NUMBER,
    4     CreationDate   DATE
    5  )
    6  PARTITION BY RANGE (CreationDate)
    7     INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
    8     (PARTITION P_FIRST
    9         VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'));
  
  Table created.
  
  SQL> INSERT INTO A VALUES (1, SYSDATE);
  
  1 row created.
  
  SQL> INSERT INTO A VALUES (1, SYSDATE - 30);
  
  1 row created.
  
  SQL> INSERT INTO A VALUES (1, SYSDATE - 60);
  
  1 row created.
  
  SQL> commit;
  
  Commit complete.
  

  You can find the form of existing partitions assistance:

  SQL> select table_name, partition_name, high_value
    2  from user_tab_partitions
    3  where table_name = 'A';
  
  TABLE_NAME PARTITION_NAME HIGH_VALUE
  ---------- -------------- --------------------------------------------------------------------------------
  A          P_FIRST        TO_DATE(' 2001-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  A          SYS_P44        TO_DATE(' 2013-01-28 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  A          SYS_P45        TO_DATE(' 2012-12-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  A          SYS_P46        TO_DATE(' 2012-11-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  

  You can then create the table a_range with apporopriate partitions. Note that you may need to create additional in a_range partitions because the partitioning interval does not create the partitions has no data for, even if that leaves 'holes' in the partitioning scheme. So, on that basis:

  SQL> CREATE TABLE A_Range (
    2     a NUMBER,
    3     CreationDate DATE)
    4  PARTITION BY RANGE (CreationDate)
    5     (partition Nov_2012 values less than (to_date('30-nov-2012', 'dd-mon-yyyy')),
    6      partition Dec_2012 values less than (to_date('31-dec-2012', 'dd-mon-yyyy')),
    7      partition Jan_2013 values less than (to_date('31-jan-2013', 'dd-mon-yyyy')),
    8      partition MAX values less than (MAXVALUE));
  
  Table created.
  

  Now, create a regular table to use in the constituencies:

  SQL> CREATE TABLE A_tmp (
    2     a              NUMBER,
    3     CreationDate   DATE);
  
  Table created.
  

  and all partitions in Exchange:

  SQL> ALTER TABLE A
    2    EXCHANGE PARTITION sys_p44
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A_Range
    2    EXCHANGE PARTITION jan_2013
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A
    2    EXCHANGE PARTITION sys_p45
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A_Range
    2    EXCHANGE PARTITION dec_2012
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A
    2    EXCHANGE PARTITION sys_p46
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A_Range
    2    EXCHANGE PARTITION nov_2012
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> select * from a;
  
  no rows selected
  
  SQL> select * from a_range;
  
           A CREATIOND
  ---------- ---------
           1 23-NOV-12
           1 23-DEC-12
           1 22-JAN-13
  

  John

• I need to change the column of the range on my partition table

  Hello
  
  I created a partition table, but I need to change column of the range "CREATED" to "PREPARED". Two of them date format, but I can't modify this table.
  
  PARTITION OF RANGE (CREATED)--> I need to change column "CREATED" as "PREPARED".
  (
  INV08 PARTITION VALUES LESS (TO_DATE('01-SEP-2010','DD-MON-YYYY')).
  INV09 PARTITION VALUES LESS (TO_DATE('01-OCT-2010','DD-MON-YYYY')).
  INV10 PARTITION VALUES LESS (TO_DATE('01-NOV-2010','DD-MON-YYYY')).
  PARTITION INV VALUES LESS THAN (MAXVALUE)
  )
  
  How can I do?
  
  Published by: user567352 on December 23, 2010 04:10

  Hello

  As far as I know, dbms_redefinition didn't even know about the partitioning:

  1. you create a new empty table that matches your need + (that is where you defined you new partition key) +.
  2. you start the names of submiting of redefinition of existing and newly created table 'interim' table
  3. you leave enough time to get the work done
  4. you have finished redefining (the name of the table are swapped)
  And voila!

  Be sure to test thorougly the process of redefinition and the performance impact that may occur until you make it to your production !
  ;-)

• Adding a Partition to a range-Hash partitioned Table

  Hello

  I created the composite partition table (RANGE-HASH).  Oracle: 11.2, OS: HP UX

  CREATE TABLE 'XYZ '. "" TABLE_XYZ.

  (

  ----

  VARCHAR2 (54) "PROFILE" DEFAULT "000000000000000000' ENABLE NON-NULL"

  NUMBER (21.7) "CREATED_ON" DEFAULT 0 ENABLE NOT NULL,.

  ----

  )

  PARTITION BY RANGE

  (

  CREATED_ON

  )

  SUBPARTITION BY HASH

  (

  VIEW PROFILE

  )

  SUBPARTITION TEMPLATE

  (

  SUBPARTITION SP1 TABLESPACE PSAPISU,

  SUBPARTITION TABLESPACE PSAPISU SP2,

  SUBPARTITION SP3 TABLESPACE PSAPISU,

  SUBPARTITION SP4 TABLESPACE PSAPISU,

  SUBPARTITION SP5 TABLESPACE PSAPISU,

  SUBPARTITION SP6 TABLESPACE PSAPISU,

  SUBPARTITION SP7 TABLESPACE PSAPISU,

  SUBPARTITION SP8 TABLESPACE PSAPISU,

  SUBPARTITION SP9 TABLESPACE PSAPISU,

  SUBPARTITION SP10 TABLESPACE PSAPISU,

  SUBPARTITION SP11 TABLESPACE PSAPISU,

  SUBPARTITION SP12 TABLESPACE PSAPISU,

  SUBPARTITION SP13 TABLESPACE PSAPISU,

  SUBPARTITION SP14 TABLESPACE PSAPISU,

  SUBPARTITION SP15 TABLESPACE PSAPISU,

  SUBPARTITION SP16 TABLESPACE PSAPISU,

  SUBPARTITION SP17 TABLESPACE PSAPISU,

  SUBPARTITION SP18 TABLESPACE PSAPISU,

  SUBPARTITION SP19 TABLESPACE PSAPISU,

  SUBPARTITION SP20 TABLESPACE PSAPISU,

  SUBPARTITION SP21 TABLESPACE PSAPISU,

  SUBPARTITION SP22 TABLESPACE PSAPISU,

  SUBPARTITION SP23 TABLESPACE PSAPISU,

  SUBPARTITION SP24 TABLESPACE PSAPISU,

  SUBPARTITION SP25 TABLESPACE PSAPISU

  )

  (

  CREATED_ON_OCP01 PARTITION VALUES LESS THAN (20090101000000).

  CREATED_ON_OCP02 PARTITION VALUES LESS THAN (20090401000000).

  CREATED_ON_OCP03 PARTITION VALUES LESS THAN (20090701000000).

  CREATED_ON_OCP04 PARTITION VALUES LESS THAN (20091001000000).

  CREATED_ON_OCP05 PARTITION VALUES LESS THAN (20100101000000).

  CREATED_ON_OCP06 PARTITION VALUES LESS THAN (20100401000000).

  CREATED_ON_OCP07 PARTITION VALUES LESS THAN (20100701000000).

  CREATED_ON_OCP08 PARTITION VALUES LESS THAN (20101001000000).

  CREATED_ON_OCP09 PARTITION VALUES LESS THAN (20110101000000).

  CREATED_ON_OCP10 PARTITION VALUES LESS THAN (20110401000000).

  CREATED_ON_OCP11 PARTITION VALUES LESS THAN (20110701000000).

  CREATED_ON_OCP12 PARTITION VALUES LESS THAN (20111001000000).

  CREATED_ON_OCP13 PARTITION VALUES LESS THAN (20120101000000).

  CREATED_ON_OCP14 PARTITION VALUES LESS THAN (20120401000000).

  CREATED_ON_OCP15 PARTITION VALUES LESS THAN (20120701000000).

  CREATED_ON_OCP16 PARTITION VALUES LESS THAN (20121001000000).

  CREATED_ON_OCP17 PARTITION VALUES LESS THAN (20130101000000).

  CREATED_ON_OCP18 PARTITION VALUES LESS THAN (20130401000000).

  CREATED_ON_OCP19 PARTITION VALUES LESS THAN (20130701000000).

  CREATED_ON_OCP20 PARTITION VALUES LESS THAN (20131001000000).

  CREATED_ON_OCP21 PARTITION VALUES LESS THAN (20140101000000).

  CREATED_ON_OCP22 PARTITION VALUES LESS THAN (20140401000000).

  CREATED_ON_OCP23 PARTITION VALUES LESS THAN (20140701000000).

  CREATED_ON_OCP24 PARTITION VALUES LESS THAN (20141001000000).

  CREATED_ON_OCP25 PARTITION VALUES LESS THAN (20150101000000).

  PARTITION CREATED_ON_OCPMAX VALUES LESS THAN (MAXVALUE)

  )

  TABLESPACE "PSAPISU" ENABLE LINE MOVEMENT;

  Now, I am trying to add the new partition but becomes under errors:

  ALTER TABLE SAPISU. ADD PARTITION VALUES LESS THAN (20150401000000) CREATED_ON_OCP26 ZTMD_300_VERS_MANAGE

  *

  ERROR on line 1:

  ORA-14074: partition bound must gather greater than that of the last partition

  ERROR on line 1:

  ORA-14018: partition bound list contains too few elements

  I tried the SYNTAX mentioned by Oracle DOC, still getting error. Partitioned table and index management

  Could if it you please let me know, what is the exact syntax that I should follow?

  See you soon

  Sameer

  CREATED_ON_OCPMAX PARTITION SPLIT THAN (MAXVALUE)

  That don't mean to 'divide' the MAXVALUE partition.

  You cannot split a partition that contains the values of A, B, C, D, MAXVALUE to MAXVALUE; that makes no sense. ERROR on line 1:

  ORA-14080: partition cannot be split along the specified high limit

  That exception is to say you can't use MAXVALUE upper limit. You must use a value that is actually IN the partition.

  Go back and look at the definition of your partitions:

  CREATED_ON_OCP25 PARTITION VALUES LESS THAN (20150101000000).

  PARTITION CREATED_ON_OCPMAX VALUES LESS THAN (MAXVALUE)

  See which ends with: "VALUES LESS THAN (MAXVALUE)? MAXVALUE is not IN the score; This is the upper limit of the partition.

  Divide certain value > '20150101000000' and LESS THAN MAXVALUE;

  So if you try to create a partition for OCP26 you can use '20160101000000'.

  CREATE TABLE PART_TEST2)
  VARCHAR2 (54) DEFAULT PROFILE "000000000000000000' ENABLE NOT NULL,
  CREATED_ON NUMBER (21.7) DEFAULT 0 ENABLE NOT NULL
  )
  PARTITION BY RANGE (CREATED_ON)
  (
  CREATED_ON_OCP24 PARTITION VALUES LESS THAN (20141001000000).
  CREATED_ON_OCP25 PARTITION VALUES LESS THAN (20150101000000).
  PARTITION CREATED_ON_OCPMAX VALUES LESS THAN (MAXVALUE)
  )

  ALTER table split partition created_on_ocpmax part_test2
  to (20160101000000) into (partition, partition CREATED_ON_OCPMAX created_on_ocp26)

• Possible to Exchange temporary table with composite range-hash partitioned table?

  Hello

  Using oracle 11.2.0.3

  We want to clean up the data in some of our existing partitioned table.

  Afetr updates check spped that is too slow for us.

  Current table is partitioned the inetrval compoiste range-hash table.  Also it is compressed - enabled for compression of basis as well

  An interval of 1 month and 1 partition by month and 4 secondary partitions in the partition of ecah.

  You want to create tenp table with the data of celan and exchange the data in this table in the 'dirty' uisng existing partitions partition exchnage.

  Is this possible?

  The plan is

  1) create temporary table containing data for 1 partition (1 month worth of data)

  (2) clean the data here

  (3) create new temporary table with these specific data which compressed and discovered partitioned with 4 secondary partitions

  (4) table 3 for swap partition dirty using partition excahnge.

  Thaks

  I think that this can be done with a combination of Exchange and Split partition partitions. Prior to Oracle 11 g, only way of redefining tables online was DBMS_REDEFINITION package. Now, you can redefine the use of partitions for Exchange & Split. Check

  http://www.Oracle-base.com/articles/Misc/partitioning-an-existing-table-using-Exchange-partition.php

  Maintenance of Partitions

  Kind regards

• Partitioning of an existing partitioned table

  Dear Experts,

  I have a problem with redefinition of partitioning of an existing partitioned table. I have a parent table separated by a virtual column by range:

  CREATE THE PARENT TABLE

  (

  PARENT_ID NUMBER NOT NULL,

  INSERT_TIMESTAMP TIMESTAMP (6) NOT NULL,

  CLOB PARENT_DATA NOT NULL,

  BATCH NUMBER,

  DATE of INSERT_DATE generated always as (TRUNC("INSERT_TIMESTAMP"))

  )

  LOB (PARENT_DATA) AS STORE NAVIGATION (ACTIVATE ONLINE STORAGE)

  PARTITION BY RANGE (INSERT_DATE)

  (

  P2015_11 PARTITION VALUES LESS THAN (TO_DATE (' 2015-12-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ACTIVATE THE ONLINE STORAGE),

  P2015_12 PARTITION VALUES LESS THAN (TO_DATE (' 2016-01-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ACTIVATE THE ONLINE STORAGE),

  P2016_01 PARTITION VALUES LESS THAN (TO_DATE (' 2016-02-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ALLOW ONLINE STORAGE,

  PARTITION P_DEFAULT VALUES LESS THAN (MAXVALUE)

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ALLOW ONLINE STORAGE

  ))

  );

  and a child table partitioned by reference:

  CREATE ACCOUNTS FROM THE TABLE. CHILD

  (

  PARENT_ID NUMBER NOT NULL,

  DATAACC DATAACC_TBL,

  CREATION_DATE DATE,

  CONSTRAINT PARENT_D_E_PRT_FK

  FOREIGN KEY (PARENT_ID)

  (PARENT_ID) REFERENCES PARENT

  ENABLE VALIDATE

  )

  NESTED TABLE, STORE DATAACC AS DATAACC_NT

  PARTITION OF REFERENCE (PARENT_D_E_PRT_FK)

  (

  P2015_11 PARTITION,

  P2015_12 PARTITION,

  P2016_01 PARTITION,

  PARTITION P_DEFAULT);

  However, now in Oracle 12 c I am able to use the reference interval partitioning. How can I redefine the partitioning of the parent table so that there can be scope-level?

  Thank you for your help.

  Best regards, Atanas.

  To change range partition to partition interval, the table should not have MAXVALUE partition... Follow the steps below to convert to partition interval

  (1) check if all data of the MAXVALUE partition (p_default) using query below

  SELECT COUNT (*) FROM parent PARTITION (p_default).

  (2) if there is no trace in the p_default partition, drop the help below

  ALTER TABLE DROP PARTITION parent p_default;

  (3) use the clause SET INTERVAL to convert the range partition partition interval as below

  ALTER TABLE parent SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'));

  I guess, you have the range partition up to until, so there should not be any folder in p_default partition.

• Partition table with null column

  Friends,

  DB: 11 GR 2

  OS: Linux

  I'm conversion from table to table of partition and try to understand the to do.

  Table has a date column that can have null values and I am partitioning table to date.

  I creates interval partitioning and also want to use MAXVALUE so that all dates go null.

  Somehow below syntax does not work, tried to read the manual, but not able to set below.

  CREATE TABLE EMP

  (DATE_ENTERED DATE default sysdate,

  ID NUMBER (10) NULL NOT ACTIVATE.

  ACTIVATE THE FIRST NAME VARCHAR2 (200) NOT NULL,

  TURN ON LAS_NAME VARHCAR2 (200) NOT NULL

  CONSTRAINT PK_ID PRIMARY KEY (ID)

  )

  PARTITION BY RANGE (DATE_ENTERED)

  INTERVAL (NUMTODSINTERVAL(1,'day'))

  (

  partition values P_NOT_USE less (to_date ('2015-01-01', 'yyyy-mm-dd'));

  values p_max_value score less than (MAXVALUE)

  );

  Receive the error message:

  ORA-14761 maxvalue partition cannot be specified for interval partitioned objects.

  Advice to solve this problem?

  I think the problem is using the function of the interval but then how can we partition a day?

  Also is there any way/script available to find all the permissions/privileges table before a fall?

  Manual: maintenance of Partitions

  The error is pretty clear to me. You can't have a partition of maxvalue for objects partitioned interval, only beach.

  There is no logical sense to have a for range partitioning's maxvalue partition, because the whole point is to have the Oracle to create new partitions automatically when you insert something above the transition point.

  SQL> desc dba_tab_privs
  Name                                      Null?    Type
  ----------------------------------------- -------- --------------
  GRANTEE                                            VARCHAR2(128)
  OWNER                                              VARCHAR2(128)
  TABLE_NAME                                        VARCHAR2(128)
  GRANTOR                                            VARCHAR2(128)
  PRIVILEGE                                          VARCHAR2(40)
  GRANTABLE                                          VARCHAR2(3)
  HIERARCHY                                          VARCHAR2(3)
  COMMON                                            VARCHAR2(3)
  TYPE                                              VARCHAR2(24)
  

• Unusable index Partitioned table

  Oracle Version: 11.2.0.2
  OS: Linux RHEL 5
  
  I'm having a problem with the index on a partitioned table interval. This is the scenario:
  
  There are 3 indices (A, B, C) on a table. A is created on two columns (1,2), B is on the column (2), C is the column (3). Table is partitioned on column 2 apart. A and B are partitioned indexes. When I drop a partition without a clause to update index, Index C will unusable. After seeing this, I gave up the C rating and create it as a local partitioned index, then all indexes are in condition of use after the drop partition statement. Is - this mandatory on a range partitioned table all indexes on the table must be locally partitioned inorder to have indexes in usable state after the drop statement of partition (without the update index clause)? Also there will be a negative impact for all indexes on the table like locally partitioned indexes?

  >
  Is - this mandatory on a range partitioned table all indexes on the table must be locally partitioned inorder to have indexes in usable state after the drop statement of partition (without the update index clause)?
  >
  Yes - it should be obvious by looking at a simple example.

  You have an OVERALL index and drop a partition but say Oracle are NOT updating the overall index.

  How this index may be usable for anything? It is not accurate. There was GARBAGE everywhere that gets in the way of Oracle find the entries of the "good". There are the index keys in the branch blocks that belong to the partition that has been deleted. Oracle cannot use these keys to determine how the blocks of branch, he needs to find.
  >
  Also there will be a negative impact for all indexes on the table like locally partitioned indexes?
  >
  Who knows? There are maybe or maybe not.

  As with many things Oracle "it depends."

  It depends on what types of queries that you ran.

  It depends on the filter predicates used for query.

  It depends on if the maintenance simple score (add/drop/split) are important to you.

  For the maintenance of the partition only all index LOCAL is the best.

  For full table scan, an index is not intended to be used.

  For a query that needs all THE records for a COMPANY_ID given in a table partitioned by DATE an OVERALL index could be better. Oracle also use 100 index LOCAL (if there are 100 partitions) to get the same data.

  It depends on.

• migration in partitioned table

  Hello world
  
  I need to migrate on a two table range-list partitioned table, one of them is partitioned list, the other is not. The largest contains 400 million lines (partitioned), the smallest contains 90 million lines. The database cannot insert all of the table at a time, so I started to migrate the tables into pieces. I have two options:
  
  1 create pieces using a date_field (IDO_ID)
  2. create the pieces with the help of a generated field (MIGR_RANK) using the list field (TAB_KOD) and a number generated randomly, which limits the lines to migrate a line no more than 1 million of partitioning
  
  My questions are the following:
  1. is it possible that the insert is slow due to the large number (about 750) list partitions? Can I use the 2. method, inserting only one partition? Or a kind of simple is enough for partitioning columns? How does the insert works in Oracle for a partitioned table? Especially in a range-list partitioned table?
  2. How can I adjust the database to migrate two tables in one step (if possible)?
  
  Thanks in advance,
  Gabor
  
  Published by: csiszgab on Sep 28, 2012 07:47

  Whenever you provide post your Oracle version 4-digit (result of SELECT * FROM V$ VERSION).
  >
  I need to migrate on a two table range-list partitioned table, one of them is partitioned list, the other is not. The largest contains 400 million lines (partitioned), the smallest contains 90 million lines. The database cannot insert all of the table at a time, so I started to migrate the tables into pieces. I have two options:
  >
  -Oracle is certainly able to process the entire table in one so why do you say otherwise?

  You can speed up insert PARALLEL aid (for the source and target tables), using the direct-path load (APPEND indicator) and NOLOGGING mode on the target table.
  >
  1. is it possible that the insert is slow due to the large number (about 750) list partitions? Can I use the 2. method, inserting only one partition? Or a kind of simple is enough for partitioning columns? How does the insert works in Oracle for a partitioned table? Especially in a range-list partitioned table?
  >
  Your "2" is not just insert in a partition. You said your target table was partitioned ' range-list' but your source is only 'list' partitioned. Therefore a partition target: data from the "one" won't a single list.

  Load using the entire table at once without (other than the partition key) index on the target table.

• Why the partitioned table cannot create?

  Why the partitioned table cannot successfully create?

  SQL> create table hr.gps_log_his
    2  (id number,
    3  name varchar2(10),
    4  time date)
    5  tablespace ts_log_his
    6  PARTITION BY RANGE (TIME)
    7  (PARTITION udp_part2009110707 VALUES LESS THAN (TO_DATE('2009110708','yyyymmddhh24')),
    8  PARTITION udp_part2009110708 VALUES LESS THAN (TO_DATE('2009110709','yyyymmddhh24')),
    9  PARTITION udp_part2009110709 VALUES LESS THAN (TO_DATE('2009110710','yyyymmddhh24')),
   10  );
  )
  *
  Error in line 10:
  ORA-14004: Missing PARTITION key word.

  Hello

  Your problem is the comma ending line 9. Delete this.

  Concerning
  Peter

• Partitioned Tables and indexes

  Hello
  
  
  I have a question on the table and index partitioning. My scenario is:
  
  Charge 2 mio records in table T once a month. Loaded records are added to existing records, and once loaded data is never changed.
  At some point, I want to delete the older recordings, so I intend to this partition table.
  
  T table looks like:
  

  create table t (id       number(10) not null  constraint t_pk primary key,
                  period   number(10) not null,
                  contract number(10) not null,
                  attr     number(10) not null);
  
  create unique index t_ux1 on t(contract,period);
  
  create index t_ix2 on t(period);

  My plan is to partition T over the period, and I'm trying to read through the concepts
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14220/partconc.htm#g471747
  
  
  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,
  
  «1. If the table partitioning column is a subset of index keys, use a local index.»
  
  "2. If the index is unique, use a global index. If this is the case, you are finished. »
  
  
  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local
  
  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?
  
  If true, what will happen when a partion fell?
  
  
  I am new in this area, so please feel the comment as you wish.
  
  
  Concerning
  Peter
  
  
  BANNER
  ----------------------------------------------------------------
  Oracle Database 10 g Enterprise Edition release 10.2.0.3.0 - 64bi
  PL/SQL version 10.2.0.3.0 - Production
  CORE Production 10.2.0.3.0
  AMT for IBM/AIX RISC System/6000: Version 10.2.0.3.0 - production
  NLSRTL Version 10.2.0.3.0 - Production

  Peter Gjelstrup wrote:

  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,

  «1. If the table partitioning column is a subset of index keys, use a local index.»

  "2. If the index is unique, use a global index. If this is the case, you are finished. »

  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local

  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?

  A partitioned index locally can only be defined as unique if the partition key is part of the columns in the index. Imagine what the database would have to do if this is not the case: in order to verify if a newly added or updated value violates the uniqueness, it will have to travel all the partitions in a serialized operation - means that no one else could do the same thing at the same time. Since he is a killer of serious scalability in terms of locking and contention, this is not allowed.

  So: Your T_UX1 index can be defined as a unique index that is local because it contains the partition key. Although the index is not prefixed ("Prefix" means that it is divided by the left side of the columns in the index) which means that there may be access patterns where all partitions should be scanned or the optimizer cannot use a method of size of effective partition according to the way the index is reached.

  Your T_PK index cannot be set as local because it must be unique (you can not use a local non-unique index in this case), but does not contain your partition key. It must be a global index. An overall index can be partitioned as well (different from the underlying table) but it doesn't have to be.

  Depends on how you access your data you have not T_IX2 index when partitioning by this key because it corresponds to the partition key and therefore could not actually be used by the mechanism of partition pruning that limit your query to the scores of individuals.

  If you have more than one MAS environment where running queries are used longer, you should be fine with the index the in general (because they could be analyzed in parallel in parallel operations), but if you have an OLTP environment, then you should avoid local no prefix indexes due to the potential problem that you need to analyze all partitions.

  Be borne in mind that with partitioning adds an important layer of complexity to other areas: in particular the options available to the optimizer and analyze cost optimizer statistics. Depends on how you access your statistical data must be maintained on several levels now (level of score and at the global level, in the case of subpartitioning may be still at this level). If your data is important and you rely on "global" level statistics (these are always the case when the optimizer at the time analysis cannot limit access to a single partition) then in the pre - 11 g databases analyze these "global" level statistics can take a lot of time and resources, since actually , you need data several times (once for the partition and even global level).

  Presenting this partitioning may mean other potential problems in terms of execution that change (not for the better sometimes) plans and how to effectively collect statistics. Note that g 11 addresses the issue of 'statistics' by introducing the so-called "extra" global statistics. Greg Rahn wrote a [blog note | http://structureddata.org/2008/07/16/oracle-11g-incremental-global-statistics-on-partitioned-tables/] on this nice feature.

  >

  If true, what will happen when a partion fell?

  Since you're already on 10g, you can specify the database to update the scores of the local index using the UPDATE of the INDEX clause, while 9i could maintain only an overall index and it is up to you to rebuild the local index partitions after the partition DDL on the table (according to the DDL operation).

  Kind regards
  Randolf

  Oracle related blog stuff:
  http://Oracle-Randolf.blogspot.com/

  SQLTools ++ for Oracle (Open source Oracle GUI for Windows):
  http://www.sqltools-plusplus.org:7676 /.
  http://sourceforge.NET/projects/SQLT-pp/

  Published by: Randolf Geist on Sep 30, 2008 16:39

  Added statistics / optimizer warning when you use the partitioning