Creating a partitioned table to an existing table interval.

Similar Questions

• Create no partition Table partitioning

  Dear all,

  I have a table that is not partition, and it has about 20 G data... If I want that table with say DATE_M partition column

  Please can anyone suggest the best way to do it.

  Thank you

  To create a partitioned table

  (1) you must make a backup of the existing one you can take expdp or DEC

  (2) remove the table and create the same table now with partitions (a minimum score is required)

  you want a range of ex-based partition if you use a date column

  (3) loading data from the previous table into the new partitioned table, either by DEC or impdp, the inserts will be automatically in their respective partitions and you don't mention explicitly.

  Their is no other possible way to partition a non partitioned table, you can do than workarounds to minimize the downtime of the table trying different ways to load data into the partitioned table.

  You can use the enable row movement clause of create table and also use local indexes for easy maintenance partition.

• Create a partitioned table

  Hi all
  I want to create a table that may exist for the next 100 years (pessimistic how I am), or maybe for the next 1000 years (now, I expanded my mind) in Oracle 9i (... oh I can't help it). I want to create a partitioned table that will store the data into partitions on annual basis and scores on monthly basis, I create a table with partitions? If Yes, then how?
  
  * 009 *.

  Yes, 9i, you create explicitly in each partition. You would normally create not all partitions in future when the table was created in the first place - you would normally a process that regularly added additional partitions.

  11.1 and later versions, you can use range partitioning to have Oracle automatically creates new partitions that they became necessary.

  Justin

• How to create a partitioned table

  Apex 4.2

  Oracle 11g

  There is a large table, which have 160000 rows, I would like to create a partitioned table to improve performance, but could not find a way to create a.

  Concerning

  Hello

  I don't see how your question relates to APEX, unless you mean how to make APEX builder object Explorer.

  Answer is simple, you can do it from the object browser.

  You should write a ddl script to create the partitioned table, and run it from sql commands.

  Kind regards

  Jari

• I install Windows 7 on a new hard drive, but the installation gives an installation message could not create a partition or locate an existing partition.

  * Original title: installation of windows 7

  I install Windows 7 on a new hard drive, but the installation gives an installation message could not create a partition or locate an existing partition. Why installation does not create a new partition.

  Try the following

  Boot from the Windows 7 DVD

  Click Install now

  Accept the license agreement

  When the option is displayed to select a type of installation, click (Custom advanced)

  Press SHIFT + F10 from your keyboard

  type the following commands

  DISKPART

  Press enter

  the list volume

  Next, you select the volume where you want to install Windows 7

  Type: select disk #.

  # represents the number of the volume

  Type: clean

  convert mbr

  create the primary partition

  assets

  format fs = ntfs quick

  output

  output

  Try to reinstall.

• Create the partition table

  Hello Experts,

  I only have 4 distinct values in my column of company. I would just create a table with partition depend on these values (201402, 201403,201404, 201405). Should I use LIST or RANGE?

  I just want the table for each distinct value of the part = 201402, 201403,201404, 201405

  So, should what partition key word I use? LIST or RANGE? or OTHERS?

  Create table MY_TABLE NOLOGGING

  as

  SELECT company_id, name...

  Of...

  WHERE THE...

  LIST BY SCORE (company_id);

  However when I use the above, that it does not work, should we specify the values? I want he's leaving for each distinct value? Even if so, should we indicate?

  Furthermore, I want to use partitioning because once I created this table, I could delete e.g. company_id = 201404, rather than delete, I can drop all of the part belonging to the 201404 does make sense? However, at the same time, I want to insert as new campanyid insert into select... where company_id = 201406;  When I insert a new campany_id the Oracle creates a new partition for it?

  Thanks in advance

  Concerning

  Charlie

  Of course you can. It would take less time to test that in order to post and wait for the reply:

  Scott@ORCL > CREATE TABLE company)
  2                       id,
  3                       name
  4                      )
  PARTITION of 5 PER LIST (id)
  6      (
  7 VALUES of company 1 PARTITION (201402),
  8 PARTITION company2 VALUES (201403).
  9 PARTITION company3 VALUES (201404)
  10)
  11 NOLOGGING
  12 AS
  13. SELECT deptno case
  14 when 10 then 201402
  15. when 20 then 201403
  16 when 30 then 201404
  end 17
  ename 18
  19 FROM emp
  20.

  Table created.

  Scott@ORCL >

  SY.

• How to create the partition on a column in a table on a daily basis

  Hi all
  I wrote a query that does not, I have a column of varchar type, and I want to create the partition of the table on that column based daily interval. but ist is in error...
  
  my query is...
  
  create the table user_detail
  (username varchar2 (50))
  The user ID number,
  country varchar2 (25).
  accupation varchar2 (100),
  company varchar2 (100),
  joindate varchar2 (14))
  partition by range (joindate)
  INTERVAL (NUMTOYMINTERVAL (1, 'DAY'))
  (PARTITION newone VALUES LESS THAN (to_date (April 13, 19 ', 'aa-mm-dd')));
  
  
  error is:
  
  Error from the 1 in the command line:
  create the table user_detail
  (username varchar2 (50))
  The user ID number,
  country varchar2 (25).
  accupation varchar2 (100),
  company varchar2 (100),
  joindate varchar2 (14))
  partition by range (joindate)
  INTERVAL (NUMTOYMINTERVAL (1, 'DAY'))
  (PARTITION newone VALUES LESS THAN (to_date (April 13, 19 ', 'aa-mm-dd')))
  Error in the command line: 1 column: 0
  Error report:
  SQL error: ORA-14751: type of invalid data for the partitioning column of a partitioned table interval
  
  
  Can I create partition on the varchar data type?

  On ORA11gR1 it works:

  --virtual column
  CREATE TABLE TEST(joindate VARCHAR2(14), part_joindate AS (to_date(joindate,'dd.mm.yyyy'))) partition by range (part_joindate)
  INTERVAL(NUMTOYMINTERVAL(1, 'YEAR'))
  (PARTITION newone VALUES LESS THAN (to_date('2013-04-19', 'yyyy-mm-dd')))
  ;
  
  INSERT INTO TEST(joindate) VALUES('19.04.2013');
  SELECT * FROM TEST;
  
  DROP TABLE TEST;
  --aux column
  CREATE TABLE TEST(joindate VARCHAR2(14), part_joindate AS (to_date(joindate,'dd.mm.yyyy'))) partition by range (part_joindate)
  INTERVAL(NUMTOYMINTERVAL(1, 'YEAR'))
  (PARTITION newone VALUES LESS THAN (to_date('2013-04-19', 'yyyy-mm-dd')))
  ;
  
  INSERT INTO TEST(joindate) VALUES('19.04.2013');
  SELECT * FROM TEST;
  
  DROP TABLE TEST;
  

  Published by: spajdy on April 19, 2013 14:58

• Conversion of non partitioned in partitioned table

  
  Hi gurus,

  I need to convert the partition table not in the partition.  More flexible way is to use the DBMS_REDEFINITION package for this.
  I do not have access to run this package, when I asked the EXECUTE permission for my dev
  CUSTOMER rejected suggestion that
  "DBMS_REDEFINITION is a method very slow migration that has never been used before here for these migrations.
  so I do not recommend using it as it can trigger bugs and unexpected side effects.

  
  is this true?

  What will be the alternative method, I can go away?

  Please suggest

  S

  I don't think DBMS_REDEFINITION has bugs.  However, you (and the client) should be familiar with the steps involved.

  Other that that, you will need to create a partitioned table and insert data to the existing table.  You can speed up integration through parallel direct-path insert.

  You will also need to build indexes on the new (partitioned) table.  Define constraints if necessary.  Run grants to other patterns, if necessary.

  Hemant K Collette

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

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

  Hello

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

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

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

• 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.

• Local primary key on reference partitioned Table

  Oracle running on Red Hat Linux Rel6 11.2.0.3.

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

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

  (entire annual <-primary key)

  whole Source_ID,

  load_dt date,

  date of doc_dt,

  doc_info clob,

  ...)

  which is partitioned on column source_id.

  I created a partitioned table of reference as follows:

  create table doc_entities_prt)

  whole doc_entity_id

  all annual,

  whole entity_id,

  forced doc_ent_fk (annual) references to documents (annual)

  tablespace...

  allow the movement of the line

  benchmark score (doc_ent_fk);

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

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

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

  Table created.

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

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

  The index is created.

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

  SQL >

  SY.

• Unusable index Partitioned table

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

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

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

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

  As with many things Oracle "it depends."

  It depends on what types of queries that you ran.

  It depends on the filter predicates used for query.

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

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

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

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

  It depends on.

• Disable LOGGING for partition table based on the automatic INTERVAL

  Hello
  
  I created a database table of paritioned range interval with NOLOGGING as the default attribute for the table as well as the tablespace definition.
  
  When ORACLE automatically creates a new partition for this table, it activates the LOGGING for this automatic partition even if the table definition says anything else.
  
  How can I go about changing this behavior? Is it still possible?
  
  FYI my platform is 11 GR 1 (11.1.0.7) material on SUN SOLARIS 10 SPARC.
  
  Any help will be appreciated.

  Right, so refer to a new feature of Oracle 11 g partitioning called 'partitioning interval. "

  +"+
  + The interval partitioning: A new strategy of partitioning in Oracle Database 11g, +.
  + Interval partitioning extends the functionality of the method range to define equipartitioned +.
  + ranges using a definition of the interval. Rather than specify individual +.
  + going to explicitly, Oracle will create any partition automatically as needed.
  + every time the data of a partition are inserted for the first time. Interval +.
  + a lot of partitioning improves maneuverability of a partitioned table. For +.
  example, + a partitioned table interval could be set so that Oracle creates a +.
  + new partition for each month in a calendar year; a partition is then automatically +.
  + created for "September 2007" as soon as the first record in this month is inserted.
  + in the database. +
  + The techniques available for a partitioned table interval are interval, interval-+.
  + List, range-Hash, and interval range. +
  +"+

  In order to disable interval partitioning on the transactions table, use:
  
  ALTER TABLE transactions SET INTERVAL ();
  

  http://download.Oracle.com/docs/CD/B28359_01/server.111/b32024/part_admin.htm

  Disable partitioning interval on this table, create a procedure to run every day we'll say, to create a new partition and chop the old partition of the appropriate table. I presume that the automatic partitioning interval creates the partition with the logging by default option without checking the parameters in the table. I have seen no information on this in the Oracle documentation.

  Hope that helps.

  Ogan

• Method of partition table

  Hello
  
  I need to partition a table with values from another table reference. I explain:
  
  I have this two tables:
  
  TABLE 1
  Identification number
  Number of ID_TABLE2
  
  
  TABLE 2
  Date to the DATE
  Identification number
  
  index (ID)
  
  I need to create a new table as TABLE1 partitioned by date, but which presents the values of TABLE2. must be something like this...
  
  CREATE TABLE newtable AS AS SELECT * FROM table1
  LIST BY SCORE (ID.)
  (
  PARTITION P1 VALUES ('Select ID from TABLA2 where DATE =' 2010-01-01').
  PARTITION P2 VALUES ('Select ID from TABLA2 where DATE =' 2010-01-01');
  )
  
  It s possible to do?
  Any sugestion another way is welcome.
  
  Kind regards

  Hello

  You cannot create a partition table based on the dynamic value. You can create it like this.

  CREATE TABLE newtable AS AS SELECT * FROM table1
  LIST BY SCORE (ID.)
  (
  PARTITION P1 VALUES ('A ', ' B', 'C').
  PARTITION P2 VALUES (WAS E ',' F ',' ');
  )

  Concerning

• ORA-14030: partitioning column does not exist in the CREATE TABLE statement

  Hi all
  
  We are trying to create a partition materialized view and get an error below.
  ORA-14030: partitioning column does not exist in the CREATE TABLE statement
  Our GL_BALANCES21 and GL_CODE_COMBINATIONS21 base tables is already divided by interval of the range on Code_combination_id.
  In the same way that we try to partition the view materialized
  We get the error.
  ORA-14030: partitioning column does not exist in the CREATE TABLE statement
  Where the clause there are 4 tables gl_balances21, gl_code_combinations21, gl_periods and gl_set_of_books.
  
  
  CREATE MATERIALIZED VIEW apps. BAL_PART
  PARTITION BY RANGE ("CODE_COMBINATION_ID")
  (SCORE LOWER (80000) VALUES,
  PARTITION OF LOWER VALUES (160000),
  PARTITION OF LOWER VALUES (240000),
  PARTITION OF LOWER VALUES (320000),
  PARTITION OF LOWER VALUES (400000),
  PARTITION OF LOWER VALUES (480000),
  PARTITION OF LOWER VALUES (560000),
  PARTITION OF LOWER VALUES (640000),
  PARTITION OF LOWER VALUES (720000),
  PARTITION OF VALUES LESS THAN (800000),
  PARTITION OF LOWER VALUES (880000),
  PARTITION OF LOWER VALUES (960000),
  PARTITION OF VALUES LESS THAN (10400000),
  PARTITION OF LOWER VALUES (11200000),
  PARTITION OF LOWER VALUES (12000000),
  PARTITION OF LOWER VALUES (12800000),
  PARTITION OF VALUES LESS THAN (13600000),
  PARTITION OF LOWER VALUES (14400000),
  PARTITION OF VALUES LESS THAN (15200000),
  PARTITION OF LOWER VALUES (16000000),
  PARTITION OF VALUES LESS THAN (16800000),
  PARTITION OF VALUES LESS THAN (17600000),
  PARTITION OF VALUES LESS THAN (18400000),
  PARTITION OF VALUES LESS THAN (19200000),
  PARTITION OF LOWER VALUES (20000000),
  PARTITION OF VALUES LESS THAN (20800000),
  PARTITION OF VALUES LESS THAN (21600000),
  PARTITION OF VALUES LESS THAN (22400000),
  PARTITION OF VALUES LESS THAN (23200000),
  PARTITION OF LOWER VALUES (24000000),
  PARTITION OF VALUES LESS THAN (24800000),
  PARTITION OF VALUES LESS THAN (25600000),
  PARTITION OF VALUES LESS THAN (26400000),
  PARTITION OF LOWER VALUES (27200000),
  PARTITION OF LOWER VALUES (28000000),
  PARTITION OF VALUES LESS THAN (28800000),
  PARTITION OF VALUES LESS THAN (29600000),
  PARTITION OF VALUES LESS THAN (30400000),
  PARTITION VALUES LESS THAN (MAXVALUE))
  QUICKLY REFRESH ON DEMAND
  SELECT the QUERY REWRITE as
  SELECT GL.GL_CODE_COMBINATIONS21. ROWID C1,
  GL.GL_BALANCES21. ROWID C2,
  "GL". "" GL_BALANCES21 ". "" ACTUAL_FLAG, "
  "GL". "" GL_BALANCES21 ". "" CURRENCY_CODE "
  "GL". "" GL_BALANCES21 ". "" PERIOD_NUM, "
  "GL". "" GL_BALANCES21 ". "" PERIOD_YEAR ".
  "GL". "" GL_BALANCES21 ". "" SET_OF_BOOKS_ID ""SOB_ID"
  "GL". "" GL_CODE_COMBINATIONS21 ". "" CODE_COMBINATION_ID ""CCID.
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT1 ",.
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT10, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" DIRECTION11, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT12, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT13, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT14, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT2 ",.
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT3. "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT4, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT5, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT6, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT7. "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT8, "
  "GL". "" GL_CODE_COMBINATIONS21 ". "" SEGMENT9, "
  "GL". "" "" GL_PERIODS '. "" PERIOD_NAME,"
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_CR', 0) Open_Bal_Cr,
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_CR', 0) +.
  NVL ("GL". "GL_BALANCES21" "." " (PERIOD_NET_CR', 0) Close_Bal_Cr,
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_DR', 0) Open_Bal_Dr,
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_DR', 0) +.
  NVL ("GL". "GL_BALANCES21" "." " (PERIOD_NET_DR', 0) Close_Bal_Dr,
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_DR', 0).
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_CR', 0) Open_Bal,
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_DR', 0).
  NVL ("GL". "GL_BALANCES21" "." " (BEGIN_BALANCE_CR', 0) +.
  NVL ("GL". "GL_BALANCES21" "." " (PERIOD_NET_DR', 0).
  NVL ("GL". "GL_BALANCES21" "." " (PERIOD_NET_CR', 0) Close_Bal,
  NVL ("GL". "GL_BALANCES21" "." " (PERIOD_NET_CR', 0) Period_Cr,
  NVL ("GL". "GL_BALANCES21" "." " (PERIOD_NET_DR', 0) Period_Dr
  OF GL.GL_CODE_COMBINATIONS21.
  GL.GL_BALANCES21,
  GL.GL_SETS_OF_BOOKS,
  GL.GL_PERIODS
  WHERE GL.GL_BALANCES21. CODE_COMBINATION_ID = GL.GL_CODE_COMBINATIONS21. CODE_COMBINATION_ID
  AND GL.GL_SETS_OF_BOOKS. SET_OF_BOOKS_ID = GL.GL_BALANCES21. SET_OF_BOOKS_ID
  AND GL.GL_PERIODS. PERIOD_NUM = GL.GL_BALANCES21. PERIOD_NUM
  AND GL.GL_PERIODS. PERIOD_YEAR = GL.GL_BALANCES21. PERIOD_YEAR
  AND GL.GL_PERIODS. PERIOD_TYPE = GL.GL_BALANCES21. PERIOD_TYPE
  AND GL.GL_PERIODS. PERIOD_NAME = GL.GL_BALANCES21. PERIOD_NAME
  AND GL.GL_PERIODS. PERIOD_SET_NAME = GL.GL_SETS_OF_BOOKS. PERIOD_SET_NAME
  and gl.GL_CODE_COMBINATIONS21.summary_flag! = « Y »
  
  ERROR on line 54:
  ORA-01013: user has requested the cancellation of the current operation
  
  I checked the metalink note saying that ensure that all columns in a partitioning column list are columns of
  the table being created.
  
  Partition is already there, on the column of code_combination_id of gl_balances21 and gl_code_combinations21.
  
  Please suggest.
  
  Thank you

  It's your mistake:

  PARTITION BY RANGE ("CODE_COMBINATION_ID") 
  

  but in your projection of column list, you have an alias he:

  "GL"."GL_CODE_COMBINATIONS21"."CODE_COMBINATION_ID" "CCID",
  

  You must use the alias as a partition key, not the name fom the secondary table column.
  --
  John Watson
  Oracle Certified Master s/n
  http://skillbuilders.com