ONLINE TABLE RANGE PARTITIONING

Hello Gurus,
I have huge table which i want to partition by RANGE(DATE) .
select count(*) from ECTMT.TEST_BORAL;
COUNT(*)
1070985
Columns are :
NID NUMBER,
TICKET_NBR VARCHAR2(20),
CHANGE_DATE DATE, (I want to partitioned by this column value )
CHANGE_DESC VARCHAR2(4000),
OPERATOR VARCHAR2(255),
TYPE VARCHAR2(255),
LAST_MODIFIED DATE
Now, the issue is space. I know the simple partioned method in which we can create the emty table and load data into it.
But, i don't want to do that. i want to do a partition into exisiting table (ECTMT.TEST_BORAL) only, in other word i can say online split partitioning.
If somebody can help,, really appreciate
Regards,
Srinivas kumar

You cannot partition a table that is not currently partitioned. You will have to create a new partitioned table and move the data from the current table into the partitioned table.
I am a little concerned that your problem is with space. You have a 1 million row table (not particularly big) and each row is no more than 5k, so we're only talking about 5 GB of data (plus another few GB, likely, for indexes). If you're trying to use partitioning in a database that doesn't have 5 GB of free space, you're probably going to have much bigger problems down the line.
You could create a new, empty, partitioned table that had a single partition that covered all CHANGE_DATE values, do a partition exchange to load the current table into your new partitioned table, and then proceed to split the one partition into the proper size and number of partitions. That is going to be slower than the alternative options, however, and involve more downtime, so it's not something that would be generally recommended. Getting 5 GB of space temporarily is likely a far easier solution.
Justin

Similar Messages

Online Table redefinition(partition maintenance) in 10g

Hi,
I would like to know more about the 10g feture of online table redifinition. I am specifically looking for removing the partition or adding new partition to the the exisitng table while the sessions are accessing the table. Can you please guide me on this.
Thanks
Anand

I recommend you writing procedure for
appending paritions
dropping partitions
from existing partition tables.
Here is simple example of dropping partitions, remember, you should drop only when parition is no longer used by application. Similarly you can write pl/sql to append partitions
DECLARE
v_msg VARCHAR2 (200);
v_table VARCHAR2 (30);
v_partiion_cur VARCHAR2 (200);
v_sql VARCHAR2 (200);
CURSOR table_cur
IS
SELECT DISTINCT table_name
FROM user_tab_partitions
Where <your condition>= criteria;
-- reteriving partition qualified for being dropped ; pass retention period
CURSOR partition_date_cur (i_table_name IN varchar2
IS
SELECT TO_DATE (TO_NUMBER (SUBSTR (partition_name, 9, 10)),
'YYYYMMDDHH24'
partition_date,
partition_name,
tablespace_name,
table_name
FROM user_tab_partitions
WHERE table_name = i_table_name
AND TO_DATE (TO_NUMBER (SUBSTR (partition_name, 9, 10)),
'YYYYMMDDHH24'
) < (SELECT TRUNC (SYSDATE)
- (SELECT VALUE
FROM some_retention_period_conf
WHERE code = 'RETENTION_PERIOD')
FROM DUAL);
BEGIN
-- some logging message
OPEN table_cur;
FETCH table_cur INTO v_table;
CLOSE table_cur;
FOR partition_date_rec IN partition_date_cur (v_table)
LOOP
v_sql :=
'ALTER TABLE '
|| partition_date_rec.table_name
|| ' DROP PARTITION '
|| partition_date_rec.partition_name;
EXECUTE IMMEDIATE v_sql;
END LOOP;
EXCEPTION
WHEN OTHERS
THEN
v_msg :=
'Partition Removal Procedure Failed : ' || SUBSTR (SQLERRM, 1, 150);
END ;

Range Partitioning a table. Max value to be defined

Hi,
I am using a range partitioned table, range partitioned on date, and have defined max value as 6 months after the Creation Date.
I have a proc which creates the partitions I want in advance by splitting up the max partition.
- Now what do I do when max partition is reached after 6 months?
- If I define max partition one year or two year after the current date instead of the currently defined 6 months after creation date. What are the negatives attached with it?
I can't use Interval Partition and have to use Range only.
Kindly suggest.
Thanks..

>
I am using a range partitioned table, range partitioned on date, and have defined max value as 6 months after the Creation Date.
I have a proc which creates the partitions I want in advance by splitting up the max partition.
- Now what do I do when max partition is reached after 6 months?
- If I define max partition one year or two year after the current date instead of the currently defined 6 months after creation date. What are the negatives attached with it?
>
Any data with a partition key that does NOT match any partition will cause your INSERT query to fail.
Any partition that has no data to match it will simply remain empty.
A common partitioning scheme is to define one partition for all old data, one partition with a high max value and then split the max value partition to get the partitions you want in the middle.
Let's say you want monthly partitions but don't have that much data from before the current year, 2012.
1. Create one partition for dates < 1/1/2012
2. One partition each for the 12 months of 2012
3. One max value partition to be 1/1/4000
You would just split the max value partition to create each month of 2013. The split could be done ahead of time or a month at a time as you choose.
The only negative is that any data inserted by mistake that has a super-high date will go into the max value partition. But that is going to happen anyway. If you accidentally enter a date of 3/23/3882 it won't be rejected.
But it is easy to query periodically to see if you have any 'bad' data like that. And the alternative is that an INSERT would fail because of the one bad record and all of your good data would be rejected anyway so it's not really much of a negative.
Remember - for best management performance each partition should have its own tablespace and the indexes should all be local if possible.

Best approach to do Range partitioning on Huge tables.

Hi All,
I am working on 11gR2 oracle 3node RAC database. below are the db details.
SQL> select * from v$version;
BANNER
Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
PL/SQL Release 11.2.0.3.0 - Production
CORE 11.2.0.3.0 Production
TNS for Linux: Version 11.2.0.3.0 - Production
NLSRTL Version 11.2.0.3.0 - Production
in my environment we have 10 big transaction (10 billion rows) tables and it is growing bigger and bigger. Now the management is planning to do a range partition based on created_dt partition key column.
We tested this partitioning startegy with few million record in other environment with below steps.
1. CREATE TABLE TRANSACTION_N
PARTITION BY RANGE ("CREATED_DT")
( PARTITION DATA1 VALUES LESS THAN (TO_DATE(' 2012-08-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') ) TABLESPACE &&TXN_TAB_PART1,
PARTITIONDATA2 VALUES LESS THAN (TO_DATE(' 2012-09-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') ) TABLESPACE &&TXN_TAB_PART2,
PARTITION DATA3 VALUES LESS THAN (TO_DATE(' 2012-10-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') ) TABLESPACE &&TXN_TAB_PART3
as (select * from TRANSACTION where 1=2);
2. exchange partion for data move to new partition table from old one.
ALTER TABLE TRANSACTION_N
EXCHANGE PARTITION DATA1
WITH TABLE TRANSACTION
WITHOUT VALIDATION;
3. create required indexes (took almost 3.5 hrs with parallel 16).
4. Rename the table names and drop the old tables.
this took around 8 hrs for one table which has 70 millions of records, then for billions of records it will take more than 8 hrs. But the problem is we get only 2 to 3 hrs of down time in production to implement these change for all tables.
Can you please suggest the best approach i can do, to copy that much big data from existing table to the newly created partitioned table and create required indexes.
Thanks,
Hari

>
in my environment we have 10 big transaction (10 billion rows) tables and it is growing bigger and bigger. Now the management is planning to do a range partition based on created_dt partition key column.
We tested this partitioning startegy with few million record in other environment with below steps.
1. CREATE TABLE TRANSACTION_N
PARTITION BY RANGE ("CREATED_DT")
( PARTITION DATA1 VALUES LESS THAN (TO_DATE(' 2012-08-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') ) TABLESPACE &&TXN_TAB_PART1,
PARTITIONDATA2 VALUES LESS THAN (TO_DATE(' 2012-09-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') ) TABLESPACE &&TXN_TAB_PART2,
PARTITION DATA3 VALUES LESS THAN (TO_DATE(' 2012-10-01 00:00:00', 'YYYY-MM-DD HH24:MI:SS') ) TABLESPACE &&TXN_TAB_PART3
as (select * from TRANSACTION where 1=2);
2. exchange partion for data move to new partition table from old one.
ALTER TABLE TRANSACTION_N
EXCHANGE PARTITION DATA1
WITH TABLE TRANSACTION
WITHOUT VALIDATION;
3. create required indexes (took almost 3.5 hrs with parallel 16).
4. Rename the table names and drop the old tables.
this took around 8 hrs for one table which has 70 millions of records, then for billions of records it will take more than 8 hrs. But the problem is we get only 2 to 3 hrs of down time in production to implement these change for all tables.
Can you please suggest the best approach i can do, to copy that much big data from existing table to the newly created partitioned table and create required indexes.
>
Sorry to tell you but that test and partitioning strategy is essentially useless and won't work for you entire table anyway. One reasone is that if you use the WITHOUT VALIDATION clause you must ensure that the data being exchanged actually belongs to the partition you are putting it in. If it doesn't you won't be able to reenable or rebuild any primary key or unique constraints that exist on the table.
See Exchanging Partitions in the VLDB and Partitioning doc
http://docs.oracle.com/cd/E18283_01/server.112/e16541/part_admin002.htm#i1107555
>
When you specify WITHOUT VALIDATION for the exchange partition operation, this is normally a fast operation because it involves only data dictionary updates. However, if the table or partitioned table involved in the exchange operation has a primary key or unique constraint enabled, then the exchange operation is performed as if WITH VALIDATION were specified to maintain the integrity of the constraints.
If you specify WITHOUT VALIDATION, then you must ensure that the data to be exchanged belongs in the partition you exchange.
>
Comments below are limited to working with ONE table only.
ISSUE #1 - ALL data will have to be moved regardless of the approach used. This should be obvious since your current data is all in one segment but each partition of a partitioned table requires its own segment. So the nut of partitioning is splitting the existing data into multiple segments almost as if you were splitting it up and inserting it into multiple tables, one table for each partition.
ISSUE#2 - You likely cannot move that much data in the 2 to 3 hours window that you have available for down time even if all you had to do was copy the existing datafiles.
ISSUE#3 - Even if you can avoid issue #2 you likely cannot rebuild ALL of the required indexes in whatever remains of the outage windows after moving the data itself.
ISSUE#4 - Unless you have conducted full volume performance testing in another environment prior to doing this in production you are taking on a tremendous amount of risk.
ISSUE#5 - Unless you have fully documented the current, actual execution plans for your most critical queries in your existing system you will have great difficulty overcoming issue #4 since you won't have the requisite plan baseline to know if the new partitioning and indexing strategies are giving you the equivalent, or better, performance.
ISSUE#6 - Things can, and will, go wrong and cause delays no matter which approach you take.
So assuming you plan to take care of issues #4 and #5 you will probably have three viable alternatives:
1. use DBMS_REDEFINITION to do the partitioning on-line. See the Oracle docs and this example from oracle-base for more info.
Redefining Tables Online - http://docs.oracle.com/cd/B28359_01/server.111/b28310/tables007.htm
Partitioning an Existing Table using DBMS_REDEFINITION
http://www.oracle-base.com/articles/misc/partitioning-an-existing-table.php
2. do the partitioning offline and hope that you don't exceed your outage window. Recover by continuing to use the existing table.
3. do the partitioning offline but remove the oldest data to minimize the amount of data that has to be worked with.
You should review all of the tables to see if you can remove older data from the current system. If you can you could use online redefinition that ignores older data. Then afterwards you can extract this old data from the old table for archiving.
If the amount of old data is substantial you can extract the new data to a new partitioned table in parallel and not deal with the old data at all.

Creating Local partitioned index on Range-Partitioned table.

Hi All,
Database Version: Oracle 8i
OS Platform: Solaris
I need to create Local-Partitioned index on a column in Range-Partitioned table having 8 million records, is there any way to perform it in fastest way.
I think we can use Nologging, Parallel, Unrecoverable options.
But while considering Undo and Redo also mainly time required to perform this activity....Which is the best method ?
Please guide me to perform it in fastest way and also online !!!
-Yasser

YasserRACDBA wrote:
3. CREATE INDEX CSB_CLIENT_CODE ON CS_BILLING (CLIENT_CODE) LOCAL
NOLOGGING PARALLEL (DEGREE 14) online;
4. Analyze the table with cascade option.
Do you think this is the only method to perform operation in fastest way? As table contain 8 million records and its production database.Yasser,
if all partitions should go to the same tablespace then you don't need to specify it for each partition.
In addition you could use the "COMPUTE STATISTICS" clause then you don't need to analyze, if you want to do it only because of the added index.
If you want to do it separately, then analyze only the index. Of course, if you want to analyze the table, too, your approach is fine.
So this is how the statement could look like:
CREATE INDEX CSB_CLIENT_CODE ON CS_BILLING (CLIENT_CODE) TABLESPACE CS_BILLING LOCAL NOLOGGING PARALLEL (DEGREE 14) ONLINE COMPUTE STATISTICS;
If this operation exceeds particular time window....can i kill the process?...What worst will happen if i kill this process?Killing an ONLINE operation is a bit of a mess... You're already quite on the edge (parallel, online, possibly compute statistics) with this statement. The ONLINE operation creates an IOT table to record the changes to the underlying table during the build operation. All these things need to be cleaned up if the operation fails or the process dies/gets killed. This cleanup is supposed to be performed by the SMON process if I remember correctly. I remember that I once ran into trouble in 8i after such an operation failed, may be I got even an ORA-00600 when I tried to access the table afterwards.
It's not unlikely that your 8.1.7.2 makes your worries with this kind of statement, so be prepared.
How much time it may take? (Just to be on safer side)The time it takes to scan the whole table (if the information can't read from another index), the sorting operation, plus writing the segment, plus any wait time due to concurrent DML / locks, plus the time to process the table that holds the changes that were done to the table while building the index.
You can try to run an EXPLAIN PLAN on your create index statement which will give you a cost indication if you're using the cost based optimizer.
Please suggest me if any other way exists to perform in fastest way.Since you will need to sort 8 million rows, if you have sufficient memory you could bump up the SORT_AREA_SIZE for your session temporarily to sort as much as possible in RAM.
-- Use e.g. 100000000 to allow a 100M SORT_AREA_SIZE
ALTER SESSION SET SORT_AREA_SIZE = <something_large>;
Regards,
Randolf
Oracle related stuff blog:
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/

Non-Partitioned Global Index on Range-Partitioned Table.

Hi All,
Is it possible to create Non-Partitioned Global Index on Range-Partitioned Table?
We have 4 indexes on CS_BILLING range-partitioned table, in which one is CBS_CLIENT_CODE(*local partitioned index*) and others are unknown types of index to me??
Means other 3 indexes are what type indexes ...either non-partitioned global index OR non-partitioned normal index??
Also if we create index as :(create index i_name on t_name(c_name)) By default it will create Global index. Please correct me......
Please help me in identifying other 3 indexes types by referring below ouputs!!!
select INDEX_NAME,TABLE_NAME,PARTITIONING_TYPE,LOCALITY from dba_part_indexes where TABLE_NAME='CS_BILLING';
INDEX_NAME TABLE_NAME PARTITI LOCALI
CSB_CLIENT_CODE CS_BILLING RANGE LOCAL
select index_name,index_type,table_name,table_type,PARTITIONED from dba_indexes where table_name='CS_BILLING';
INDEX_NAME INDEX_TYPE TABLE_NAME TABLE_TYPE PAR
CSB_CREATE_DATE NORMAL CS_BILLING TABLE NO
CSB_SUBMIT_ORDER NORMAL CS_BILLING TABLE NO
CSB_CLIENT_CODE NORMAL CS_BILLING TABLE YES
CSB_ORDER_NBR NORMAL CS_BILLING TABLE NO
select INDEX_OWNER,INDEX_NAME,TABLE_NAME,COLUMN_NAME from dba_ind_columns where TABLE_NAME='CS_BILLING';
INDEX_OWNER INDEX_NAME TABLE_NAME COLUMN_NAME
RPADMIN CSB_CREATE_DATE CS_BILLING CREATE_DATE
RPADMIN CSB_SUBMIT_ORDER CS_BILLING SUBMIT_TO_INVOICE
RPADMIN CSB_SUBMIT_ORDER CS_BILLING ORDER_NBR
RPADMIN CSB_CLIENT_CODE CS_BILLING CLIENT_CODE
RPADMIN CSB_ORDER_NBR CS_BILLING ORDER_NBR
select dip.index_name, dpi.locality, dip.partition_name, dip.status
from dba_part_indexes dpi, dba_ind_partitions dip
where dpi.table_name ='CS_BILLING'
and dpi.index_name = dip.index_name;
INDEX_NAME LOCALI PARTITION_NAME STATUS
CSB_CLIENT_CODE LOCAL CSB_2006_4Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2006_3Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2007_1Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2007_2Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2007_3Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2007_4Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2008_1Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2008_2Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2008_3Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2008_4Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2009_1Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2009_2Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2009_3Q USABLE
CSB_CLIENT_CODE LOCAL CSB_2009_4Q USABLE
select * from dba_part_indexes
where table_name ='CS_BILLING'
and locality = 'GLOBAL';
no rows selected
-Yasser
Edited by: YasserRACDBA on Mar 5, 2009 11:45 PM

Yaseer,
Is it possible to create Non-Partitioned and Global Index on Range-Partitioned Table?
Yes
We have 4 indexes on CS_BILLING range-partitioned table, in which one is CBS_CLIENT_CODE(*local partitioned index*) and others are unknown types of index to me??
Means other 3 indexes are what type indexes ...either non-partitioned global index OR non-partitioned normal index??
You got local index and 3 non-partitioned "NORMAL" b-tree tyep indexes
Also if we create index as :(create index i_name on t_name(c_name)) By default it will create Global index. Please correct me......
Above staement will create non-partitioned index
Here is an example of creating global partitioned indexes
CREATE INDEX month_ix ON sales(sales_month)
   GLOBAL PARTITION BY RANGE(sales_month)
      (PARTITION pm1_ix VALUES LESS THAN (2)
       PARTITION pm2_ix VALUES LESS THAN (3)
       PARTITION pm3_ix VALUES LESS THAN (4)
        PARTITION pm12_ix VALUES LESS THAN (MAXVALUE));Regards

Problem with a 2 columns Range Partitioning for a indexed organized table

have an indexed organized table with a 2 column PK. the first field (datum) is a date field the second field (installatieid) is a number(2) field.
Every minute a 7 records are inserted (installatieid 0-6).
I like to partition this table with one partition per year per installatieid.
I tried to do it with:
partition by range(datum,installatieid)
(partition P_2004_0 values less than (to_date('2004-01,'yyyy-mm'),1)
,partition P_2004_6 values less than (to_date('2004-01','yyyy-mm'),7)
partition P_2005_0 values less than (to_date('2005-01','yyyy-mm'),1)
,partition P_2005_6 values less than (to_date('2005-01','yyyy-mm'),7)
but now only the P_2004_0 and P_2005_0 are filled.
I thought about to combine a range partition on datum with a list subpartition on installatieid, but I read this is not allowed with an index organized table.
How can I solve this problem.

partition by range(datum,installatieid)
(partition P_2004_0 values less than
(to_date('2004-01,'yyyy-mm'))
,partition P_2004_6 values less than
(to_date('2004-07','yyyy-mm'))
partition P_2005_0 values less than
(to_date('2005-01','yyyy-mm'))
,partition P_2005_6 values less than
(to_date('2005-07','yyyy-mm'))
? Sorry haven't got time to test it this morning ;0)

Partition Pruning on Interval Range Partitioned Table not happening when SYSDATE used in Where Clause

We have tables that are interval range partitioned on a DATE column, with a partition for each day - all very standard and straight out of Oracle doc.
A 3rd party application queries the tables to find number of rows based on date range that is on the column used for the partition key.
This application uses date range specified relative to current date - i.e. for last two days would be "..startdate > SYSDATE -2 " - but partition pruning does not take place and the explain plan shows that every partition is included.
By presenting the query using the date in a variable partition pruning does table place, and query obviously performs much better.
DB is 11.2.0.3 on RHEL6, and default parameters set - i.e. nothing changed that would influence optimizer behavior to something unusual.
I can't work out why this would be so. It very easy to reproduce with simple test case below.
I'd be very interested to hear any thoughts on why it is this way and whether anything can be done to permit the partition pruning to work with a query including SYSDATE as it would be difficult to get the application code changed.
Furthermore to make a case to change the code I would need an explanation of why querying using SYSDATE is not good practice, and I don't know of any such information.
1) Create 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) Populate table 1million rows spread between 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) Query the Table for data from last 2 days using SYSDATE in clause
EXPLAIN PLAN FOR
SELECT count(*)
FROM    part_test
WHERE   starttime >= SYSDATE - 2;
| Id | Operation                 | Name      | Rows | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
|   0 | SELECT STATEMENT          |           |     1 |     8 | 7895 (1)| 00:00:01 |       |       |
|   1 | SORT AGGREGATE           |           |     1 |     8 |            |          |       |       |
|   2 |   PARTITION RANGE 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|
Predicate Information (identified by operation id):
   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 answer but very different cost.
EXPLAIN PLAN FOR
SELECT count(*)
FROM part_test
WHERE starttime >= (to_date('23122013:0950','DDMMYYYY:HH24MI'))-2;
| Id | Operation                 | Name      | Rows | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
|   0 | SELECT STATEMENT          |           |     1 |     8 |   131 (0)| 00:00:01 |       |       |
|   1 | SORT AGGREGATE           |           |     1 |     8 |            |          |       |       |
|   2 |   PARTITION RANGE 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|
Predicate Information (identified by operation id):
   3 - filter("STARTTIME">=TO_DATE(' 2013-12-21 09:50:00', 'syyyy-mm-dd hh24:mi:ss'))
thanks in anticipation
Jim

As Jonathan has already pointed out there are situations where the CBO knows that partition pruning will occur but is unable to identify those partitions at parse time. The CBO will then use a dynamic pruning which means determine the partitions to eliminate dynamically at run time. This is why you see the KEY information instead of a known partition number. This is to occur mainly when you compare a function to your partition key i.e. where partition_key = function. And SYSDATE is a function. For the other bizarre PSTOP number (1048575) see this blog
http://hourim.wordpress.com/2013/11/08/interval-partitioning-and-pstop-in-execution-plan/
Best regards
Mohamed Houri

Unique key on range-partitioned table

Hi,
We are using a composite range-hash interval partitioned table
Uses index - trying to make this have same tablespace as the partitions i.e. local but not liking it
alter table RETAILER_TRANSACTION_COMP_POR
add constraint RETAILER_TRANSACTION_COMP_PK primary key (DWH_NUM)
using index
LOCAL
ora-14039: partitioning columns must form a subset of key columns of a unique index
Without local then fine but doesn't have same tablespace as the partitions and don't want to make this part of partition key.
Tbale range partitioned - this is just a UK to prevent 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

Modify HUGE HASH partition table to RANGE partition and HASH subpartition

I have a table with 130,000,000 rows hash partitioned as below
----RANGE PARTITION--
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY RANGE (YRMO_NBR)(
PARTITION TEST_PART_200009 VALUES LESS THAN(200009),
PARTITION TEST_PART_200010 VALUES LESS THAN(200010),
PARTITION TEST_PART_200011 VALUES LESS THAN(200011),
PARTITION TEST_PART_MAX VALUES LESS THAN(MAXVALUE)
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR, LINE_ID);
Data: -
INSERT INTO TEST_PART
VALUES ('2000',200001,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200009,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200010,'CM');
VALUES ('2006',NULL,'CM');
COMMIT;
Now, I need to keep this table from growing by deleting records that fall b/w a specific range of YRMO_NBR. I think it will be easy if I create a range partition on YRMO_NBR field and then create the current hash partition as a sub-partition.
How do I change the current partition of the table from HASH partition to RANGE partition and a sub-partition (HASH) without losing the data and existing indexes?
The table after restructuring should look like the one below
COMPOSIT PARTITION-- RANGE PARTITION & HASH SUBPARTITION --
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY RANGE (YRMO_NBR)
SUBPARTITION BY HASH (C_NBR) (
PARTITION TEST_PART_200009 VALUES LESS THAN(200009) SUBPARTITIONS 2,
PARTITION TEST_PART_200010 VALUES LESS THAN(200010) SUBPARTITIONS 2,
PARTITION TEST_PART_200011 VALUES LESS THAN(200011) SUBPARTITIONS 2,
PARTITION TEST_PART_MAX VALUES LESS THAN(MAXVALUE) SUBPARTITIONS 2
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR,LINE_ID);
Pls advice
Thanks in advance

Sorry for the confusion in the first part where I had given a RANGE PARTITION instead of HASH partition. Pls read as follows;
I have a table with 130,000,000 rows hash partitioned as below
----HASH PARTITION--
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY HASH (C_NBR)
PARTITIONS 2
STORE IN (PCRD_MBR_MR_02, PCRD_MBR_MR_01);
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR,LINE_ID);
Data: -
INSERT INTO TEST_PART
VALUES ('2000',200001,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200009,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200010,'CM');
VALUES ('2006',NULL,'CM');
COMMIT;
Now, I need to keep this table from growing by deleting records that fall b/w a specific range of YRMO_NBR. I think it will be easy if I create a range partition on YRMO_NBR field and then create the current hash partition as a sub-partition.
How do I change the current partition of the table from hash partition to range partition and a sub-partition (hash) without losing the data and existing indexes?
The table after restructuring should look like the one below
COMPOSIT PARTITION-- RANGE PARTITION & HASH SUBPARTITION --
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY RANGE (YRMO_NBR)
SUBPARTITION BY HASH (C_NBR) (
PARTITION TEST_PART_200009 VALUES LESS THAN(200009) SUBPARTITIONS 2,
PARTITION TEST_PART_200010 VALUES LESS THAN(200010) SUBPARTITIONS 2,
PARTITION TEST_PART_200011 VALUES LESS THAN(200011) SUBPARTITIONS 2,
PARTITION TEST_PART_MAX VALUES LESS THAN(MAXVALUE) SUBPARTITIONS 2
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR,LINE_ID);
Pls advice
Thanks in advance

How to see the logic in an existing range partition table of a database.

Hi I need to see the logic used in a Range partitioned table in a database.
I need this because i need to create a db similar to this.
This is a live database.
Version used :10.2.0.3.0
How can i seee this
Cheers,
Kunwar

How to get that.
For example if i want to see the script for a table. I get this eror
SQL> select dbms_metadata.get_ddl('TABLE','x) from dual;
ERROR:
ORA-31603: object "x" of type TABLE not found in schema "SYS"
ORA-06512: at "SYS.DBMS_METADATA", line 1546
ORA-06512: at "SYS.DBMS_METADATA", line 1583
ORA-06512: at "SYS.DBMS_METADATA", line 1901
ORA-06512: at "SYS.DBMS_METADATA", line 2792
ORA-06512: at "SYS.DBMS_METADATA", line 4333
ORA-06512: at line 1
And i dont know the login details of that user..

Range Partitioning Existing table

Hi,
I have a table with numeric field. I want to range partition the existing table. I used the following query
alter table dept add partition by range(deptno) (
partition p1 values less than (20),
partition p1 values less than (40))
It gives me ORA-00902: invalid datatype.
What am I doing wrong? Is it possible to partition an existing table?
Regards,
Subbu S.

Hi,
You cannot partition an existing table with an ALTER TABLE command. There are couple of techinques with which you can partition an existing table, like:
1) Rename existing table, create a new partitoned table and move data.
2) Using DBMS_REDEFINITION package.
Have a look at the following links for details:
http://www.oracle-base.com/articles/misc/PartitioningAnExistingTable.php
http://www.dba-oracle.com/oracle_news/2004_3_9_rittman.htm
Regards
Asif Momen
http://momendba.blogspot.com

Range Partitioning of Table

Hi all,
I need to do range partitioning on a table using current system timestamp. sth like
CREATE TABLE SAMPLE(ID NUMBER, SAMPLE_DATE TIMESTAMP)PARTITION BY RANGE(SAMPLE_DATE)(PARTITION P1 VALUES LESS THAN (LOCALTIMESTAMP));
Is there any way I can assign a dynamic date using variable other than string literal after VALUES LESS THAN clause? I tried achieving this with stored procedure but always throws error for this filed.
CREATE OR REPLACE PROCEDURE partition_test
IS
x TIMESTAMP;
comm long;
BEGIN
SELECT LOCALTIMESTAMP into x from dual;
comm := 'CREATE TABLE SAMPLE(ID NUMBER, SAMPLE_DATE TIMESTAMP)PARTITION BY RANGE(SAMPLE_DATE)(PARTITION P1 VALUES LESS THAN (x))';
execute immediate comm;
END partition_test;
error I'm getting is
ORA-14019: partition bound element must be one of: string, datetime or interval literal, number, or MAXVALUE
Any clue? Would really appreciate your help.
Noman

Can you please try with proper quotes like this...
comm := 'CREATE TABLE SAMPLE(ID NUMBER, SAMPLE_DATE TIMESTAMP)PARTITION BY RANGE(SAMPLE_DATE)(PARTITION P1 VALUES LESS THAN (' || x || '))';
If that didn't work out, try printing the value of comm before execute immediate and if required you can add proper number of quotes.

Create list-range partition table

Database version: Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
I am trying to create a partition table with LIST-Range, and I am getting this following error, is oracle 10.2.0.4 database supports to create list-range partition(composite)
SQL> CREATE TABLE tbp_list_range
REPORT_DATE DATE,
member_id_01 varchar2(2),
DATE_SERVICE date,
member_id varchar2(15)
PARTITION BY LIST(member_id_01)
SUBPARTITION BY RANGE (DATE_SERVICE)
PARTITION SPTYR04M01_C VALUES('AA','aa')
NOLOGGING
COMPRESS (
SUBPARTITION PTYR12M011 VALUES LESS THAN (TO_DATE(' 2012-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN')),
SUBPARTITION PTYR12M021 VALUES LESS THAN (TO_DATE(' 2012-02-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN')),
SUBPARTITION recent1 VALUES LESS THAN (MAXVALUE)
PARTITION SPTYR04M01_Yo VALUES('BJ','bj')
NOLOGGING
COMPRESS (
SUBPARTITION PTYR12M01 VALUES LESS THAN (TO_DATE(' 2012-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN')),
SUBPARTITION PTYR12M02 VALUES LESS THAN (TO_DATE(' 2012-02-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN')),
SUBPARTITION recent2 VALUES LESS THAN (MAXVALUE)
SUBPARTITION BY RANGE (DATE_SERVICE)
ERROR at line 9:
ORA-00922: missing or invalid option
SQL> select * from v$version;
BANNER
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
PL/SQL Release 10.2.0.4.0 - Production
CORE 10.2.0.4.0 Production
TNS for Linux: Version 10.2.0.4.0 - Production
NLSRTL Version 10.2.0.4.0 - Production
Any help would be greatly appreciated.
Thanks,

any possible work around for this scenario on 10.2.0.4
--thanks

TIMESTAMP(6) Partitioned Key - Range partitioned table ddl needed

What is DDL syntax for TIMESTAMP(6) Partitioned Key, Range partitioned table
Edited by: oracletune on Jan 11, 2013 10:26 AM

>
What is DDL syntax for TIMESTAMP(6) Partitioned Key, Range partitioned table
>
Not sure what you are asking. Are you asking how to create a partitioned table using a TIMESTAMP(6) column for the key?
CREATE TABLE TEST1
    USERID                 NUMBER,
    ENTRYCREATEDDATE     TIMESTAMP(6)
PARTITION BY RANGE (ENTRYCREATEDDATE) INTERVAL(NUMTOYMINTERVAL(1, 'MONTH'))
    PARTITION P0 VALUES LESS THAN (TO_DATE('1-1-2013', 'DD-MM-YYYY'))
)See my reply Posted: Jan 10, 2013 9:56 PM if you need to do it on a TIMESTAMP with TIME ZONE column. You need to add a virtual column.
Creating range paritions automatically

ONLINE TABLE RANGE PARTITIONING

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