Range of partitioning of the interval

Similar Questions

• Query for the partitions of the interval, system and reference

  Hi, I want to ask some of range partitions, system and reference information.

  In the partition of the interval, the script looks like this

  CREATE TABLE interval_sales

  (prod_id NUMBER (6))

  cust_id NUMBER

  time_id DATE

  )

  PARTITION OF RANGE (time_id)

  INTERVAL (NUMTOYMINTERVAL (1, 'MONTH'))

  (PARTITION p0 VALUES LESS THAN (TO_DATE (' 01/01/2005 ',' DD-MM-YYYY "")));

  I would like to write a query to find the expression of the interval "MONTHS" but could not find what he makes.

  In the reference partition, the script looks like this

  CREATE TABLE child_tab)

  ID NUMBER NOT NULL,

  parent_tab_id NUMBER NOT NULL,

  code VARCHAR2 (10),

  Description VARCHAR2 (50).

  CREATED_DATE DATE,

  CONSTRAINT child_tab_pk PRIMARY KEY (id),

  CONSTRAINT child_parent_tab_fk FOREIGN KEY (parent_tab_id)

  REFERENCES parent_tab (id)

  )

  PARTITION BY REFERENCE (child_parent_tab_fk);

  I want to write a query to find the foreign key name 'child_parent_tab_fk' after the PARTITION BY REFERENCE to the last line, but have no luck so far.

  In the end, I want to achieve this goal:

  If (interval_partitioning_type)

  Assign the interval expression to a string variable

  Another yew (reference_partitioning_type)

  Set the foreign key to a variable of type of foreign key

  Is there a way to get the type of partitioning? (interval and reference here)

  Please advise,

  Jack

  You can query ALL_PART_TABLES

  REF_PTN_CONSTRAINT_NAME	VARCHAR2(30)		Name of the referential partitioning constraint for partitioned tables reference
  INTERVAL	VARCHAR2(1000)		The value of the interval chain

• Partitions of the interval

  Hi guys,.

  I use DBMS_REDEFINITION to partition tables not partitioned in order to facilitate better performance. My question is; These tables already have clues about them. I read scores interval may not work on nested index tables. I need advice here. Can I used partitions of the interval on tables with indexes on the?

  by default, oracle creates a table as an array of heap and stores the rows in its blocks without a special order. On a heap table (which can be partitioned), you can create indexes. An index organized table (IOT) is basically an index without an additional table of the segment (but you can also define a segment of overflow, but it is another story). There are some cases of use of IOT, but in most situations, a heap table is quite sufficient.

  By the way: here's an example for the redefinition of a table with partitions of the interval: http://uhesse.com/2010/02/15/partitioning-a-table-online-with-dbms_redefinition/

• Partition wise joined possible with partitions of the interval?

  Hello

  I want to know the score wise join (NTC) is possible with interval partitioning - I can't find an explicit statement that he isn't, but I can't make it work - I did a simple test case to illustrate the issue.

  below, I have 2 create table scripts - 1 for the case of interval and 1 for the case of hash - I then a simple query on these 2 objects which should produce a NTC.

  In the case of hash, it works very well (see screenshot 2nd with a set of slaves), the first screenshot shows the case of the interval where I find myself with 2 sets of slaves and no NTC.

  No idea if this is possible and I just missed something?

  (for the test case choose the names of schema/storage appropriate for your system)

  Oh and version (I almost forgot... :-))-East 11.2.0.4.1 SLES 11)

  See you soon,.

  Rich

  -case interval

  CREATE TABLE 'SB_DWH_IN '. "' TEST1 '.

  TABLESPACE "SB_DWH_INTEGRATION".

  PARTITION BY RANGE ("OBJECT_ID") INTERVAL (10000)

  (PARTITION 'LESS_THAN_ZERO' VALUES LESS THAN (0) TABLESPACE "SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  CREATE TABLE 'SB_DWH_IN '. "" TEST2 ".

  TABLESPACE "SB_DWH_INTEGRATION".

  PARTITION BY RANGE ("OBJECT_ID") INTERVAL (10000)

  (PARTITION 'LESS_THAN_ZERO' VALUES LESS THAN (0) TABLESPACE "SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  -case of hash

  CREATE TABLE 'SB_DWH_IN '. "' TEST1 '.

  TABLESPACE "SB_DWH_INTEGRATION".

  8 partitions PARTITION OF HASH ("OBJECT_ID")

  store in ("SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  CREATE TABLE 'SB_DWH_IN '. "" TEST2 ".

  TABLESPACE "SB_DWH_INTEGRATION".

  8 partitions PARTITION OF HASH ("OBJECT_ID")

  store in ("SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  -query to run

  Select / * + PARALLEL(TEST2,8) PARALLEL(TEST1,8) * / *.

  of 'SB_DWH_IN '. "" TEST2 ","SB_DWH_IN ". "' TEST1 '.

  where TEST1.object_id = test2.object_id

  

  

  It is planned and a consequence of the estimate of the number of parallel slaves.

  To the parallel 41 each slave made 3 passes (i.e. sleeves 3 partitions).

  Add a partition (by table), and a set of slaves will have to manage a 4th pass: the cost of the query using NTC would increase from 33 percent even if the modification of the data is less than 0.8%.

  I guess that in the production Oracle distributes your lines of 1 M for a hash join.

  Because the decision is encrypted, it is possible that a very extreme tilt in partition in the table sizes billion line might overthrow the optimizer in a non - NTC join - but I have not tested that.

  If you want to force the plan John Watson suggestion for a hint of pq_distribute is relevant.  To cover all the bases and call your tables SMALL and LARGE

  /*+

  leading (FAT kid)

  USE_HASH (large)

  no_swap_join_inputs (large)

  PQ_DISTRIBUTE (wide none none)

  */

  If it's legal, that should do it.

  Concerning

  Jonathan Lewis

• How to search the dictionary of data for the partitions of the interval?

  Oracle 11.2.0.3

  When I have a DBA_PART_TABLES request. PARTITIONING_TYPE he says "beach". I want to be able to differential an interval of other partitions of range based in the data dictionary?

  
  used this example. Watch "range".

  http://www.rittmanmead.com/2008/09/investigating-Oracle-11g-interval-partitioning/

  Oracle Docs list 'SYSTEM' as a partitioning_type that I thought would be intervals.

  ALL_PART_TABLES

  [code]

  1 CREATE TABLE interval_sales

  2 (prod_id NUMBER (6))

  3, cust_id NUMBER

  4, time_id DATE

  5, channel_id char (1)

  6, promo_id NUMBER (6)

  7, quantity_sold NUMBER (3)

  8, amount_sold NUMBER (10.2)

  9          )

  10 PARTITION BY RANGE (time_id)

  INTERVAL OF 11 (NUMTOYMINTERVAL (1, 'MONTH'))

  12 (PARTITION p0 VALUES LESS THAN (TO_DATE (' 01/01/2005 ',' DD-MM-YYYY "")),)

  13 PARTITION p1 VALUES LESS THAN (TO_DATE (' 01/01/2006 ',' DD-MM-YYYY ""));

  14 p2 PARTITION VALUES LESS THAN (TO_DATE (' 07/01/2006 ',' DD-MM-YYYY ""));

  15 * PARTITION p3 VALUES LESS THAN (TO_DATE (' 01/01/2007 ',' DD-MM-YYYY "")))

  SQL > /.

  Table created.

  Elapsed time: 00:00:00.01

  SQL > select partitioning_type from the user_part_tables where table_name = 'INTERVAL_SALES ';

  PARTITION

  ---------

  RANGE

  [/ code]

  CREATE TABLE junk1
     ( id         NUMBER )
     PARTITION BY RANGE (id)
     INTERVAL(5)
       ( PARTITION p0 VALUES LESS THAN (5),
         PARTITION p1 VALUES LESS THAN (10),
         PARTITION p2 VALUES LESS THAN (15),
         PARTITION p3 VALUES LESS THAN (20) )
  /
  
  CREATE TABLE junk2
     ( id         NUMBER )
     PARTITION BY RANGE (id)
       ( PARTITION p0 VALUES LESS THAN (5),
         PARTITION p1 VALUES LESS THAN (10),
         PARTITION p2 VALUES LESS THAN (15),
         PARTITION p3 VALUES LESS THAN (20) )
  /
  
  select owner, table_name, partitioning_type, interval
    from dba_part_tables
   where table_name in ('JUNK1', 'JUNK2')
  /
  

  TABLE_NAME OWNER PARTITIONING_TYPE INTERVAL

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

  GREGS                          JUNK1                          RANGE             5

  GREGS JUNK2 RANGE

  2 selected lines.

• overall index using partitioning of the interval.

  Can we have an index that is partitioned across the world using range partitioning on the number field in the database table.

  >
  Can we have an index that is partitioned across the world using range partitioning on the number field in the database table.
  >
  No - a global index cannot be paritioned of the interval.
  You can create a range partitioned or Hash-Partitioned Global index, but it cannot use partitioning interval. The index may be on a table is partitioned apart, but the index itself cannot be partitioned apart.

  See Creation of Partitions in the VLDB and partitioning Guide database
  http://docs.Oracle.com/CD/E18283_01/server.112/e16541/part_admin001.htm#i1006455

  and the CREATE INDEX SQL language doc section
  http://docs.Oracle.com/CD/E11882_01/server.112/e26088/statements_5012.htm#i2062403

• Interval partitioning using the key of the virtual column

  Hi guys

  I have a series of tables that I propose to the partition.

  Each table has a year and a period (equivalent to a month) and I intend using those as my partition keys - year as the partition and the period as the subpartition.

  The distribution will be actually:

  2015

  001

  002

  ...

  012

  2016

  001

  002

  ...

  012

  etc...

  I would use range partitioning for the year, because it would make the maintenance of the partition a breeze.  Only one problem - the field is of type varchar, and is not modifiable as its legacy applications that have become bigger than Ben Hurr and more tangled than a Ben Hurr size flat spaghetti.

  It is hence the idea of virtual column between in game - I could create a virtual column in these tables to convert the field to digital and then partition on it instead.

  My question-if the year of origin varchar column is included in the where clause of a query, is the optimizer based on CSSTidy smart enough to realize that the virtual column used for the partition key is based on this column and use pruning of partition to improve performance?  Or queries must refer to the virtual column to make it work?  Also, the virtual column should be indexed to make it work?

  Thanks in advance.

  
  

  Simon
  

  orclz >

  orclz > create table pt (v1 varchar2 (1) c1 as (to_number (v1)), number of c2)

  2 partition by range (c1) interval (1)

  subpartition by hash (c2) subpartitions 4 3

  4 (partition p1 values less than (1))

  5.

  Table created.

  orclz > set autot on explain

  orclz > select * PT where v1 = '0';

  no selected line

  Execution plan

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

  Hash value of plan: 711571056

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

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

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

  |  0 | SELECT STATEMENT |      |    82.  2296 |    29 (0) | 00:00:01 |      |      |

  |  1.  RANGE OF PARTITION ALL THE |      |    82.  2296 |    29 (0) | 00:00:01 |    1. 1048575.

  |  2.  HASH PARTITION ALL |      |    82.  2296 |    29 (0) | 00:00:01 |    1.    4.

  |*  3 |    TABLE ACCESS FULL | PT |    82.  2296 |    29 (0) | 00:00:01 |    1. 1048575.

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

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

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

  3 - filter ("V1" = '0')

  orclz > select * PT where c1 = 0;

  no selected line

  Execution plan

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

  Hash value of plan: 1726115854

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

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

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

  |  0 | SELECT STATEMENT |      |    82.  2296 |    29 (0) | 00:00:01 |      |      |

  |  1.  RANGE OF SINGLE PARTITION |      |    82.  2296 |    29 (0) | 00:00:01 |    1.    1.

  |  2.  HASH PARTITION ALL |      |    82.  2296 |    29 (0) | 00:00:01 |    1.    4.

  |*  3 |    TABLE ACCESS FULL | PT |    82.  2296 |    29 (0) | 00:00:01 |    1.    4.

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

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

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

  3 - filter ("C1" = 0)

  --

  John Watson

  Oracle Certified Master s/n

• Impact of setting the interval partitioning off and then on again

  Hi all

  Have discovered that to split the partitions of a partitioned table Beach-interval, in that we need to define the partitioning interval all issue the command split then on again.

  e, g.

  ALTER TABLE retailer_transaction SET INTERVAL ();

  ALTER TABLE POR2_retailer_transaction

  SPLIT PARTITION PART_400012 AT (TO_DATE (' 2020-09-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))

  INTO (PARTITION PART_202008,

  PART_400012 PARTITION)

  ALTER TABLE retailer_transaction SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'))

  Everything we need to pay attention to that?

  Using oracle 11.2.0.3

  Thank you

  You can NOT split the LAST (most HIGH_VALUE) partition of a partitioned table interval interval. You can only separate partitions BEFORE the last. Which converts also all partitions from the bottom (split and below) for the partitions of the RANGE.

  If interval partitioning will only apply to data added above the transition point (the score higher.

  To return to your original question. To split the LAST interval follow the procedure described in your original post

  1. range partitioning

  2 split the partitioning

  3 set the partitioning interval on

  Conclusion with this however when adding data back to a nine month doesn't automatically create the partition again.

  This "new month" MUST BE above the point of passage for a new partition to create. In your code example, this point of transition is for the year 4000.

  This works very well for me:

  ALTER TABLE sales PARTITION RENAME SYS_P1122 to SALE_400012

  ALTER TABLE sale SET INTERVAL ();

  ALTER TABLE SPLIT PARTITION for sale

  SALE_400012 TO (TO_DATE('01/02/2020','DD/MM/YYYY')) IN

  (

  SALE_202001 PARTITION,

  SALE_400012 PARTITION

  );

  ALTER TABLE sale SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'))

  INSERT INTO sales

  (SALES_DATE, STORE_NUM, QTY_SOLD)

  values (to_date('31/12/4001','DD/,MM/YYYY'), 1, 10);

  This last INSERT statement above creates a NEW partition because the date (31/12/4001) is above the transition point.

  Looks like you want to move the transition point downwards. That is - you were using this year there date of 4000 as a 'catch all '.

  1. set the partitioning interval

  2 split that last partition into two NEW partitions (new names) by using a VALUE that is high enough to contain ALL the data you have in this last partition.

  3. remove the last partition - it should now be empty

  4. turn on range partitioning

  Now, any data that you insert above this NEW point of transition (the limit that you used for the new partition) it will create a new partition of the interval. Make sure the border is really an end of month, if that's what you want.

  If your data sample had this

  INSERT INTO sales

  (SALES_DATE, STORE_NUM, QTY_SOLD)

  values (to_date('05/10/2016','DD/,MM/YYYY'), 1, 10);

  INSERT INTO sales

  (SALES_DATE, STORE_NUM, QTY_SOLD)

  values (to_date('31/12/4000','DD/,MM/YYYY'), 1, 10);

  So for the step #2 above use split date 06/2016 so that the line above for 10/05/2106 goes to the new partition and 12/31/4000 stays in the old.

  Then, drag this old and lines in it (31/12/4000).

  You can NOT split the last partition if interval is always on because it is the last partition interval.

  And as long as you have given for 12/31/4000, you will get no new partitions unless you insert was more TOP than this.

  So if you want partitioning interval to activate again 06/2016 that higher data would be deleted and the score he falls.

  You have discovered only among "traps" with the help of partitioning of the interval. Oracle will happily create new partition needed to keep the data. If ANYONE, even accidentally, between a date as 12/31/4000 that nobody would ever know until the new data seem to DISAPPEAR for some use cases.

  This is because all data will start to enter this new partition ACCIDENTALLY created instead of a partition for the month, it's part.

  And, and you have discovered, you cannot split this partition directly.

  Delayed segment creation can allow you to pre-create the partitions ahead without actually affecting the segment for them until the first row is inserted.

  Interval partitioning will create the appropriate partitions but not if someone accidentally or intentionally, data will be a valuable capitalized in the table.

• 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

• How to create the INTERVAL partitioning using EM?

  Hi guys

  I'm still not able to use the INTERVAL partitioning using EM. Is it possible to do?

  Oracle 11 g 2 on Linux

  Thank you

  John

  Hi John,.

  I don't think it's possible, that the option of partition of the range seems to have things that make you think of it would generate the definition as the partitioning interval but it doesn't.

  I'm on 12.1.0.3 of the SGC and 12.1.0.5 DB plugin so I'm sure it's 'missing' features for the moment.

  See you soon,.

  Rich

• Partition of the range

  Hello
  
  I want to ask if I can create the partition of the range of like 10 minutes? As I have data of 1 hour and I want to make a partition of 10? Please suggest
  
  Thank you

  Try place range partitions in good condition

  create table part_test ( id date) partition by range(id)
   ( partition p1 values less than(to_date('01-JAN-2012 01:00:00','DD-MON-YYYY hh24:mi:ss'))
    ,partition p2 values less than(to_date('01-JAN-2012 02:00:00','DD-MON-YYYY hh24:mi:ss'))
  .....
    ,partition p24 values less than(to_date('02-JAN-2012 00:00:00','DD-MON-YYYY hh24:mi:ss'))
   );
  

• Rebuild the range-Hash partitioned local index

  Hi all
  
  We have a table with 400 million records in our dataware House environment. It is partitioned using range-Hash Partitioning composit. He used a global Indexes previously. Last month the global indices have been changed to the index the. Since then, Exchange and index rebuild partition takes twice against the take used previously. From my understanding local index should provide availability high and more usable in the warehouse environment. Please correct me if I'm wrong and give me your suggestion. Also, I'm looking for a good material for the exchange of the partition and Index rebild with locally manage the range-hash partitioned index. Let me know if you know a good link for this...
  
  
  Thanks in advance...

  Dear krmreddy,

  Carefully read the following link and I hope this will guide you;

  http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:223318100346512896

  Kind regards.

  Ogan

• On the partition of the range

  Dear friends,
  
  I use Oracle10g. I need to create a range partition as below:
  
  
  CREATE TABLE EMP
  (
  EMPNO NUMBER 4,
  ENAME VARCHAR2 (10 BYTE),
  USE VARCHAR2 (BYTE 9),
  MGR NUMBER 4,
  HIREDATE DATE,
  SAL NUMBER (7.2).
  NUMBER (7.2), COMM.
  DEPTNO NUMBER (2)
  )
  
  PARTITION OF RANGE (hiredate)
  (
  PARTITION emp_1980 VALUES LESS THAN (TO_DATE('31/12/1980','DD/MM/YYYY')) TABLESPACE USERS,
  PARTITION emp_1981 VALUES LESS THAN (TO_DATE('31/12/1981','DD/MM/YYYY')) TABLESPACE USERS,
  PARTITION emp_1982 VALUES LESS THAN (TO_DATE('31/12/1982','DD/MM/YYYY')) TABLESPACE USERS,
  PARTITION emp_1987 VALUES LESS THAN (TO_DATE('31/12/1987','DD/MM/YYYY')) TABLESPACE USERS
  );
  
  But the problem is:
  
  (1) If no data entry is not match with the criteria above, then where he's going? I mean, not the data that don't follow the rule above which partition it will? is it possible to put the option DEFAULT partition in the partition of the range?
  I find that the DEFAULT option in the partition from the LIST
  
  (2) except that, when I add a new partition with what above, then it shows the error below:
  
  change the emp table add partition emp.emp_1988 VALUES LESS THAN (TO_DATE('31/12/1988','DD/MM/YYYY')) USERS TABLESPACE
  
  
  Error:
  "ORA-14020: this physical attribute may not be specified for the partition table."
  
  
  
  Someone could help me please, how can I solve the above problems?

  Range partitioning provides a range of values. Therefore, any value greater than the maximum value in your definitions of partition would fail. Thus, for example, 01/01/1988 and would especially error. That's why the range partitioning allows you to specify a 'best' possible values where you say 'lower values (MAXVALUE) '. Therefore, if you had included a partition "emp_maxvalue" by 'lower values (MAXVALUE)' after the partition of 'emp_1987', any value of 1988 or higher would go in the emp_maxvalue partition.

  {Because there is no minvalue specified, a value of, say, 06/01/1979 would go in the score of "emp_1980" as it is "(01/01/1980) below"!}

  Yet once, as the partitioning is a date range, you cannot Add a partition. You need to DIVIDE a partition. For example, if you had a present "emp_maxvalue" partition, in order to create the partition of emp_1988, you would be SPLIT "emp_maxvalue" into two partitions: emp_1988 and emp_maxvalue.

  Please read the documentation on the Partition maintenance in the Administrator's Guide.

  Hemant K Collette
  http://hemantoracledba.blogspot.com

• Look for a partition in the range

  Hi gurus,
  
  I have a partition table,
  
  CREATE TABLE sales
  (
  Number of SalesID
  SalesNumber varchar2 (50).
  Date of SalesDate,
  Number of value TotalAmount
  )
  PARTITION BY RANGE (SalesDate)
  (PARTITION P_JAN VALUES LESS THAN (TO_DATE (1 JANUARY 2008 ',' DD/MM/YYYY "")),)
  PARTITION P_FEB VALUES LESS THAN (TO_DATE (JANUARY 2, 2008 ',' DD/MM/YYYY "")).
  PARTITION P_MAR VALUES LESS THAN (TO_DATE (JANUARY 3, 2008 ',' DD/MM/YYYY "")).
  PARTITION P_APRVALUES LESS (TO_DATE (JANUARY 4, 2008 ',' DD/MM/YYYY ')),
  PARTITION P_MAX VALUES LESS THAN (MAXVALUE))
  ;
  
  I need to dynamically manage partitions based on the criteria of data,
  
  for example. to TO_DATE (January 5, 2008 ',' DD/MM/YYYY '), first I have to check if there is no partition exisits who, if so have need create this partition, a parition/month
  
  an entry for this condition is just a date, it can be any date.
  
  Please help me get the right query.
  
  Thank you

  Could you just do new lines fall under the partition of "p_max' by default and launch something to divide every so often?

  User_tab_partitions (perhaps with user_part_key_columns to check key columns) questioning made delicate by the fact that high_value is a LONG column. It may be easier to read in a PL/SQL variable and check it on the procedural level.

  11g can automatically create missing partitions on the fly so it might be easier to upgrade rather than developing your own.

• Need to create the Partition in the history table

  Hello

  I have an audit table that stores the data for the last 8 years in it and the customer wants to retain the data for the last 8 years. Now as adding data to this table is increasing, there is a delay in execution of report which are acquired by querying on this table.

  Basically, they go a from date and to date. Then these date are compared with two columns of dates in the table. The condition of the query will be as below

  TRUNC (NVL (DATE1, DATE2)) > = TRUNC (TO_DATE (?, ' dd/mm/yyyy ""))

  AND TRUNC (NVL (DATE1, DATE2)) < = TRUNC (TO_DATE (?, ' dd/mm/yyyy ""))

  Users can query for any date range.

  I am thinking of creating a Partition of the range on this table for each month during the 8 years of data. Please suggest on the question of whether I should use partition interval range to make it more effective. And would also create an Index on this partition for the above two day columns.

  12 (months) SO * 8 (years) = 98 would be created partitions. This will not overload the schema of database/Tablespace? Y at - it another effective way to set the table.

  Please suggest and let me know for any clarification.

  There are 2 approaches (but not limited to 2) to create the Partition on an existing table:

  Note: There is no ALTER TABLE statement to convert a heap table to a partitioned table.

  1. create a temporary Partitioned table with the same types of data in the column. Take a dump of the original table (HEAP) export and import the dump into the newly created partitioned intermediate table. After checking the data file the original and rename the interim even as the original table.

  2 using the package Oracle DBMS_REDEFINITION automates the steps above from said (not exactly the same steps, however) and its done online - which means there is no required downtime, the original table may undergo DML operations during this process of redefinition. View partitioning of a table online with DBMS_REDEFINITION | Oracle instructor

  According to my understanding if let us know as a partition interval then in April if there is an inserted record then it will create a new partition for the month of April will be of value less than 30 April 2014 "and all other documents created in April will pass under it"

  Yes, it's a feature of 11g, which Oracle manages adding partitions as needed. In earlier versions DBA can add partitions manually from time to time. Reference www.gavinsoorma.com/2009/09/11g-interval-partitioning/

  Here the reason for check the sent_out_datetime NVL will be NULL for few records, so in this case we want to get the records for dates of mapping them to received_datetime

  In addition, try to replace the NVL with COALESCE and test it. This would help improve performance.