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/

Tags: Database

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

  • Range of partitioning of the interval

    Hello


    Using Oracle 11.20.3
    We want to have a table partitioned composite range-hash, a partition for each month.

    However, want to esnure are still a tehrfore partition would use range interval partitioning.

    Have a few questions, if I say 201301 to 201312 but get intellectualization rosw partition with a value date in June 2014 it create
    a new partition for each month IE 201401, 201402 etc. or just a 201406?

    Wanyt esnure system can automatically dial with a scenario

    Daty1 have partitions 201301 to 201312

    Day 2 get record for 201406 (June 2014)

    Day 3 get record for 201403 (March 2014)


    We have no control over the date range, we were able to get but don't want to tpo have to manually create partitions BOF tehse all dates in advance - want system she managed automatically.

    Also you can just rename it the names of system partition generated to something more useful later.

    Yes, the creation of partitions of the interval is 'rare' - a given partition will be created only when necessary. (This should be easy to try, by the way).

  • 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

    nonPWJ.PNG

    pwjenabled.PNG

    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

  • 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

  • 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.

  • 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.

  • 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

  • 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

  • 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

  • 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.

  • importing into a partitioned table of interval 11g

    as I took export utility simple partition table 8i exp not rained so 100 k lines in there.

    and imported with the import utility in the interval of 11g partitioned based on the date column.

    There were imported, but did not what I expected...

    If we execute the simple insert for partition interval 11g command, it create new partition automatically according to the strategy of partition.

    Here's the demo...

    created range partitioned table on the date with shift interval column...

    CREATE TABLE TEST.xxx_HIST
    (
    xxx_DATE DATE NOT NULL,
    P_ROLL_CONVENTION CHAR (2),
    R_ROLL_CONVENTION CHAR (2),
    P_COMPOUNDING_IND CHAR (2),
    R_COMPOUNDING_IND CHAR (2),
    P_CALC_METHOD CHAR (2),
    R_CALC_METHOD CHAR (2),
    P_SPREAD_AMT NUMBER (28,12).
    R_SPREAD_AMT NUMBER (28,12).

    )
    partition by range (xxx_DATE)
    interval (numtoyminterval(3,'MONTH'))
    store (security)
    (
    values of pQ1 lower partition (to_date('2010-01-01','yyyy-mm-dd'))
    ) IN PARALLEL.


    -IMPORTED FROM ROWS IN THE TABLE...

    ======================================================================
    Connected to: Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - 64 bit Production
    With partitioning, OLAP, Data Mining and Real Application Testing options

    Export file created by EXPORT: V08.01.07 direct

    CAUTION: objects have been exported by SYSTEM, not by you

    . import of xx_ARCH in TEST objects
    . . import of 141749 lines imported from the table 'xxx_HIST '.
    Import completed successfully without warnings.
    ========================================================================



    -HE HAS A LOT OF DATES OF DIFF IN THERE...



    SQL > SELECT COUNT (DISTINCT xxx_DATE) TEST.xxx_HIST;

    COUNT (DISTINCT xxx_DATE)
    -----------------------------
    1371


    28-MARCH 06
    10 FEBRUARY 06
    9 FEBRUARY 05
    20 FEBRUARY 02
    3 JUNE 02
    10 MAY 04
    26 DECEMBER 03
    31 JANUARY 03

    xxx
    ---------
    21 JULY 08
    31 OCTOBER 05
    25 APRIL 08
    28 APRIL 08
    12 OCTOBER 06
    DECEMBER 21 07
    28 DECEMBER 04


    -BUT STILL ALL DUMPED INTO A PARTITION


    SQL > SELECT nom_partition FROM DBA_TAB_PARTITIONS WHERE TABLE_OWNER = 'TEST ';

    NOM_PARTITION
    ------------------------------
    PQ1

    It all dumped in a partition...

    fact partition interval 11g creates the partition automatically in function whose lines if imported... when we import lines in there...? or am I missing something?

    any idea guys?

    Seems to be a poor strategy for me because if I am not mistaken, there is no way to specify the order of the imported lines. If you import a line with the date max as your first row... bang, you get a range partition created for you and the rest falling.

    I think you'd be better import these data into a table in step and then by a

    insert into new_fancy_partition_table
select *
from old_8_temporary_imported_table
order by date_column asc

    Or create the partitions manually.

    I just realized that you specify a partition in your create table statement (missed that on cursory inspection). And I think you misunderstand how the interval works... it's for values LARGER than the existing partitions ONLY...

    http://download.Oracle.com/docs/CD/E11882_01/server.112/e10592/statements_7002.htm#SQLRF01402

    "
    INTERVAL clause

    Use this clause to set the interval of partitioning the table. Range partitions are partitions based on a digital range interval or datetime. * They extend from range partitioning by commanding the database to automatically create partitions of the specified range or interval when the data inserted in the table exceed all the partitions.* range
    "

    Published by: Tubby on August 16, 2010 18:32

    Additional document link.

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    I'm not an experienced user, but I'll do my best to explain the problem.

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    I also read that you can create new partition if you have "unallocated" space (partition), but all 4 before indicated are those primary and not affected.

    Then, the next opportunity that I read on is to do a right-click on C and "reduce volume", and then create new partition from the rest of the space (if I understand it is good), but then I read in Windows Help:
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    Please advise...

    Posted by just
    [So the next opportunity I haven't read is to do a right-click on C and "reduce volume", and then create new partition from the rest of the space (if I understand it right), FONTS] [/ COLOR]

    Yes, this is the right way to have more than one partition. You can define what capacity should partition C and any free space can be "convert" to the D partition.

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    I've had my laptop for 2 months now and when I click on computer to watch my hard drive there are two; 1 called vista (: c) and other data (: E).
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    According to my experience, the Vista operating system to the second partition for data backup. It s necessary and important if you want to restore or repair the Windows operating system. This is a new feature in Vista!

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