Adding a column to a partitioned table.

Similar Questions

• Change the Data Type of column on large partitioned Table Tablet

  Hello

  I was tasked to change a certain NUMBER of collar to a colonel VARCHAR2

  It is usually an easy task, but I ran into several challenges due to the size of the table, the second major problem is the compression of the table.

  To begin with, the Table and index are 4.4 to and hold around 11 lines billion. That the Table is compressed, not the index.

  The first option was to add a new collar of VCHAR, update from the Col de NUM, file and then drop the colonel NUM

  
  

  We ran this test in the Test environment and because discovered Compression Table that you can only mark the column is not used, and it can only be moved permanently from the Table if you unzip the whole Table and run «...» REMOVE the UNUSED COLUMNS'.

  The second way is to create a separate Table copy with the values of the source of neck, then write to null to the source of neck, change data Type when its empty and then update the source again with the copy Table values.

  The second option, that's what I'm testing, but I work with nearly 30 million lines by daily score, inserts take very long to fill.

  How can I make faster inserts between the Source and the copy Table?

  What other channels follow to accomplish the task above?

  Concerning

  Stephan

  Nowhere in your proposed solution was necessary for you to understand why the data source changes, its as simple that to take what is ons the table (all the facts in the original post) and offering the best way on the basis of previous experience or oracle skills acquired over time.

  If you spent so much time and thought to help with a solution that you did on WHY complain, you could have provided some feasible solutions already.

  No thanks to your information irrelevant and unnecessary, I've implemented a solution already.

  1 export the Source Table

  2. create a 'replica Table' using the DDL source using a temporary table name

  3. import all Partitions in the Table new

  4. before the Cut-Over, synchronize the changes from the Source to the replica Table

  5 rename the Source

  5 rename the Table of replica for the Original Source Table name

  6. slide the original Source Table

• Help please SQL... Adding a column of total accumulated in a table

  Tonight... I have the task of adding a column of total cumulative to a table that contains a list of transactions. The table contains entries for several accounts.
  
  for example
  
  create the table adam_test1
  (
  account char (4),
  date of entry_date,
  amount number 4,
  Balance number 4
  );
  
  
  insert into adam_test1 values ('1000', January 1, 2000 ", 50", null);
  insert into adam_test1 values ('1000', January 2, 2000 ', 40', null);
  insert into adam_test1 values ('1000', 3 January 2000', '-50', null ");
  insert into adam_test1 values ('1000', January 4, 2000 ', 10', null);
  insert into adam_test1 values ('1000', 5 January 2000', '-250', null ");
  insert into adam_test1 values ('1000', January 6, 2000', '100', null);
  insert into adam_test1 values ('1000', January 7, 2000 ", 50", null);
  
  insert into adam_test1 values ('2000,' 1 January 2000 ', 30', null);
  insert into adam_test1 values ('2000', January 2, 2000 ', 10', null);
  insert into adam_test1 values ('2000', 3 January 2000', '-520', null ");
  insert into adam_test1 values ('2000', January 4, 2000', '140', null);
  insert into adam_test1 values ('2000', 5 January 2000', '-4', null ");
  insert into adam_test1 values ('2000', January 6, 2000 "," 120", null);
  insert into adam_test1 values ('2000', January 7, 2000 ', 57', null);
  
  
  SQL > select * from adam_test1;
  
  ACCO ENTRY_DAT BALANCE AMOUNT
  ---- --------- ---------- ----------
  1000 1 JANUARY 00 50
  1000 2 JANUARY 00 40
  1000 3 JANUARY 00-50
  1000 4 JANUARY 00 10
  1000 5 JANUARY 00-250
  1000 6 JANUARY 00 100
  1000 7 JANUARY 00 50
  2000 1 JANUARY 00 30
  2000 2 JANUARY 00 10
  2000 3 JANUARY 00-520
  2000 4 JANUARY 00 140
  2000 JANUARY 5, 00-4
  2000-6 JANUARY 00 120
  2000-7 JANUARY 00 57
  
  .. I tried a few things to fill the values using rowid, but I was not able to get anything to work - ideas?
  Thank you!
  Adam

  Use MERGE:

  merge
    into  adam_test1 t1
    using (
           select  rowid,
                   sum(amount) over(partition by account order by entry_date) balance
             from  adam_test1
          ) t2
    on (t1.rowid = t2.rowid)
    when matched
      then
        update set t1.balance = t2.balance
  /
  
  ACCO ENTRY_DAT     AMOUNT    BALANCE
  ---- --------- ---------- ----------
  1000 01-JAN-00         50         50
  1000 02-JAN-00         40         90
  1000 03-JAN-00        -50         40
  1000 04-JAN-00         10         50
  1000 05-JAN-00       -250       -200
  1000 06-JAN-00        100       -100
  1000 07-JAN-00         50        -50
  2000 01-JAN-00         30         30
  2000 02-JAN-00         10         40
  2000 03-JAN-00       -520       -480
  2000 04-JAN-00        140       -340
  
  ACCO ENTRY_DAT     AMOUNT    BALANCE
  ---- --------- ---------- ----------
  2000 05-JAN-00         -4       -344
  2000 06-JAN-00        120       -224
  2000 07-JAN-00         57       -167
  
  14 rows selected.
  
  SQL> 
  

  SY.

• Partition table with null column

  Friends,

  DB: 11 GR 2

  OS: Linux

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

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

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

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

  CREATE TABLE EMP

  (DATE_ENTERED DATE default sysdate,

  ID NUMBER (10) NULL NOT ACTIVATE.

  ACTIVATE THE FIRST NAME VARCHAR2 (200) NOT NULL,

  TURN ON LAS_NAME VARHCAR2 (200) NOT NULL

  CONSTRAINT PK_ID PRIMARY KEY (ID)

  )

  PARTITION BY RANGE (DATE_ENTERED)

  INTERVAL (NUMTODSINTERVAL(1,'day'))

  (

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

  values p_max_value score less than (MAXVALUE)

  );

  Receive the error message:

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

  Advice to solve this problem?

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

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

  Manual: maintenance of Partitions

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

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

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

• I have a table of the adf, I added a column that contains a button that I created, when I click it must remove this row in the table, but it is not, please help

  I have a table of the adf, I added a column that contains a button that I created, when I click it must remove this row in the table, but it is not, please help

  I don't understand. You use vo and eo for you to use business components.

  Again, this kind of code call in trouble.

  You must post the changes to make them visible to the eo find vo. You must then run the query for the changes in the business layer strips then you must update the iterator he table is based on.

  In your code I see that happen, hooch maybe because it is more often than not formatted and undocumented.

  My advice is to do a small test case that you can manage with easy sql. Once you get it to run transfer you the results to the actual application.

  Timo

• Can we have hash partition and list on 2 columns in the same table. How.

  Can we have hash partition and list on 2 columns in the same table. How.
  
  Work on 10g release 2

  # List-Hash is a valid composite partitioning strategy 11 from g.

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

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

  Hello

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

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

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

• Adding a Partition to a range-Hash partitioned Table

  Hello

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

  CREATE TABLE 'XYZ '. "" TABLE_XYZ.

  (

  ----

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

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

  ----

  )

  PARTITION BY RANGE

  (

  CREATED_ON

  )

  SUBPARTITION BY HASH

  (

  VIEW PROFILE

  )

  SUBPARTITION TEMPLATE

  (

  SUBPARTITION SP1 TABLESPACE PSAPISU,

  SUBPARTITION TABLESPACE PSAPISU SP2,

  SUBPARTITION SP3 TABLESPACE PSAPISU,

  SUBPARTITION SP4 TABLESPACE PSAPISU,

  SUBPARTITION SP5 TABLESPACE PSAPISU,

  SUBPARTITION SP6 TABLESPACE PSAPISU,

  SUBPARTITION SP7 TABLESPACE PSAPISU,

  SUBPARTITION SP8 TABLESPACE PSAPISU,

  SUBPARTITION SP9 TABLESPACE PSAPISU,

  SUBPARTITION SP10 TABLESPACE PSAPISU,

  SUBPARTITION SP11 TABLESPACE PSAPISU,

  SUBPARTITION SP12 TABLESPACE PSAPISU,

  SUBPARTITION SP13 TABLESPACE PSAPISU,

  SUBPARTITION SP14 TABLESPACE PSAPISU,

  SUBPARTITION SP15 TABLESPACE PSAPISU,

  SUBPARTITION SP16 TABLESPACE PSAPISU,

  SUBPARTITION SP17 TABLESPACE PSAPISU,

  SUBPARTITION SP18 TABLESPACE PSAPISU,

  SUBPARTITION SP19 TABLESPACE PSAPISU,

  SUBPARTITION SP20 TABLESPACE PSAPISU,

  SUBPARTITION SP21 TABLESPACE PSAPISU,

  SUBPARTITION SP22 TABLESPACE PSAPISU,

  SUBPARTITION SP23 TABLESPACE PSAPISU,

  SUBPARTITION SP24 TABLESPACE PSAPISU,

  SUBPARTITION SP25 TABLESPACE PSAPISU

  )

  (

  CREATED_ON_OCP01 PARTITION VALUES LESS THAN (20090101000000).

  CREATED_ON_OCP02 PARTITION VALUES LESS THAN (20090401000000).

  CREATED_ON_OCP03 PARTITION VALUES LESS THAN (20090701000000).

  CREATED_ON_OCP04 PARTITION VALUES LESS THAN (20091001000000).

  CREATED_ON_OCP05 PARTITION VALUES LESS THAN (20100101000000).

  CREATED_ON_OCP06 PARTITION VALUES LESS THAN (20100401000000).

  CREATED_ON_OCP07 PARTITION VALUES LESS THAN (20100701000000).

  CREATED_ON_OCP08 PARTITION VALUES LESS THAN (20101001000000).

  CREATED_ON_OCP09 PARTITION VALUES LESS THAN (20110101000000).

  CREATED_ON_OCP10 PARTITION VALUES LESS THAN (20110401000000).

  CREATED_ON_OCP11 PARTITION VALUES LESS THAN (20110701000000).

  CREATED_ON_OCP12 PARTITION VALUES LESS THAN (20111001000000).

  CREATED_ON_OCP13 PARTITION VALUES LESS THAN (20120101000000).

  CREATED_ON_OCP14 PARTITION VALUES LESS THAN (20120401000000).

  CREATED_ON_OCP15 PARTITION VALUES LESS THAN (20120701000000).

  CREATED_ON_OCP16 PARTITION VALUES LESS THAN (20121001000000).

  CREATED_ON_OCP17 PARTITION VALUES LESS THAN (20130101000000).

  CREATED_ON_OCP18 PARTITION VALUES LESS THAN (20130401000000).

  CREATED_ON_OCP19 PARTITION VALUES LESS THAN (20130701000000).

  CREATED_ON_OCP20 PARTITION VALUES LESS THAN (20131001000000).

  CREATED_ON_OCP21 PARTITION VALUES LESS THAN (20140101000000).

  CREATED_ON_OCP22 PARTITION VALUES LESS THAN (20140401000000).

  CREATED_ON_OCP23 PARTITION VALUES LESS THAN (20140701000000).

  CREATED_ON_OCP24 PARTITION VALUES LESS THAN (20141001000000).

  CREATED_ON_OCP25 PARTITION VALUES LESS THAN (20150101000000).

  PARTITION CREATED_ON_OCPMAX VALUES LESS THAN (MAXVALUE)

  )

  TABLESPACE "PSAPISU" ENABLE LINE MOVEMENT;

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

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

  *

  ERROR on line 1:

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

  ERROR on line 1:

  ORA-14018: partition bound list contains too few elements

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

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

  See you soon

  Sameer

  CREATED_ON_OCPMAX PARTITION SPLIT THAN (MAXVALUE)

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

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

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

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

  Go back and look at the definition of your partitions:

  CREATED_ON_OCP25 PARTITION VALUES LESS THAN (20150101000000).

  PARTITION CREATED_ON_OCPMAX VALUES LESS THAN (MAXVALUE)

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

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

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

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

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

• Adding Partition for a partitioned Table

  Hello
  I have a partitioned table with 7 existing Partitions that follow this structure:
  
  PARTITION 'TRANSACTIONS_JAN2012' VALUES (TO_DATE () LOWER
  "2012-02-01 00:00:00 ', ' SYYYY-MM-JJ HH24:MI:SS»(, ' NLS_CALENDAR = GRÉGORIEN ')"
  ) PCTFREE, PCTUSED 10 0 INITRANS 1 MAXTRANS 255 NOLOGGING (INITIAL STORAGE
  1073741824 THEN 1048576 MINEXTENTS 1 MAXEXTENTS 2147483645 PCTINCREASE 0
  FREELISTS 1 FREELIST GROUPS 1 USER_TABLES DEFAULT FLASH_CACHE
  DEFAULT CELL_FLASH_CACHE) TABLESPACE "FTSV1" NOCOMPRESS
  
  I am trying to add a new Partition:
  ALTER table movements Add partition
  (partition TRANSACTIONS_JTS VALUES LESS THAN (TO_DATE (' 2012-03-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN ')));
  
  But get this error:
  SQL error: ORA-00902: invalid data type
  00902 00000 - "invalid data type".
  
  Any help is greatly appreciated.
  
  Thanks in advance,
  Johnny

  Hi Johnny,.

  Looks like you have one to several "Partition", delete this as well as the unnecessary parentheses.

  ALTER TABLE transactions
      ADD PARTITION transactions_jts
      VALUES LESS THAN
          (TO_DATE(' 2012-03-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'));
  

  Concerning
  Peter

• Adding a new partition to a partitioned table

  Hi all
  
  I try to add a new partition to an existing table in compartmentalized but I need the new partition of 'table' to go to tablespace DATA_2011 and partition "index" to go to INDEX_2011. I tried but it wrong on what follows.
  
  CREATE TABLE employees
  (employee_id number 4 NOT NULL,
  last_name VARCHAR2 (10),
  department_id NUMBER (2))
  PARTITION OF RANGE (department_id)
  (PARTITION employees_part1 VALUES LESS THAN (11) TABLESPACE DATA_2010,)
  PARTITION employees_part2 VALUES LESS THAN (21) TABLESPACE DATA_2010;
  PARTITION employees_part3 VALUES LESS THAN (31) TABLESPACE DATA_2010);
  
  CREATE INDEX employees_local_idx ON employees (employee_id) LOCAL
  (PARTITION TABLESPACE INDEX_2010 employees_part1,
  PARTITION employees_part2 TABLESPACE INDEX_2010,
  PARTITION employees_part3 TABLESPACE INDEX_2010);
  
  
  Here's what I'm trying to run but it fails...
  ALTER table employees add PARTITION "EMPLOYEES_PART4" VALUES LESS THAN (41) DATA_2011 index tablespace tablespace INDEX_2011;
  
  ERROR on line 1:
  ORA-14020: this physical attribute may not be specified for the partition table
  
  
  
  Succeeds what follows, but in this case the index partition "EMPLOYEES_PART4" is also assigned to DATA_2011 tablespace.
  
  ALTER table employees add PARTITION "EMPLOYEES_PART4" VALUES LESS THAN (41) tablespace DATA_2011;
  
  
  Any ideas... I am currently looking in the documentation to see if it would be possible?
  
  Thanks for your time...

  Hello

  ALTER table employees add PARTITION "EMPLOYEES_PART4" VALUES LESS THAN (41) tablespace DATA_2011;

  The foregoing is the corect order and therefore succeeded.

  but in this case the index partition "EMPLOYEES_PART4" is also assigned to DATA_2011 tablespace.

  You need rebuild the index of the partition with the name desired tablespace. /

  ALTER index rebuild partition tablespace

  Anand

• Partitioned Tables and indexes

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

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

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

  Peter Gjelstrup wrote:

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

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

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

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

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

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

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

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

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

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

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

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

  >

  If true, what will happen when a partion fell?

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

  Kind regards
  Randolf

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

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

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

  Added statistics / optimizer warning when you use the partitioning

• Partitioning of an existing partitioned table

  Dear Experts,

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

  CREATE THE PARENT TABLE

  (

  PARENT_ID NUMBER NOT NULL,

  INSERT_TIMESTAMP TIMESTAMP (6) NOT NULL,

  CLOB PARENT_DATA NOT NULL,

  BATCH NUMBER,

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

  )

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

  PARTITION BY RANGE (INSERT_DATE)

  (

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

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ACTIVATE THE ONLINE STORAGE),

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

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ACTIVATE THE ONLINE STORAGE),

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

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ALLOW ONLINE STORAGE,

  PARTITION P_DEFAULT VALUES LESS THAN (MAXVALUE)

  CRAFT STORE (PARENT_DATA) AS (SECUREFILE

  ALLOW ONLINE STORAGE

  ))

  );

  and a child table partitioned by reference:

  CREATE ACCOUNTS FROM THE TABLE. CHILD

  (

  PARENT_ID NUMBER NOT NULL,

  DATAACC DATAACC_TBL,

  CREATION_DATE DATE,

  CONSTRAINT PARENT_D_E_PRT_FK

  FOREIGN KEY (PARENT_ID)

  (PARENT_ID) REFERENCES PARENT

  ENABLE VALIDATE

  )

  NESTED TABLE, STORE DATAACC AS DATAACC_NT

  PARTITION OF REFERENCE (PARENT_D_E_PRT_FK)

  (

  P2015_11 PARTITION,

  P2015_12 PARTITION,

  P2016_01 PARTITION,

  PARTITION P_DEFAULT);

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

  Thank you for your help.

  Best regards, Atanas.

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

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

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

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

  ALTER TABLE DROP PARTITION parent p_default;

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

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

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

• 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.

• Delete the data in partitioned tables

  Hello

  Oracle 11.2.0.3.0 version, running on Linux Enterprise.

  I need to remove all the data from two tables (several 100 million lines each) that are partitioned.

  (1) table 1 is a partition table have varied to a DATE column

  (2) Table2 is a partition of reference table; partitioning is referenced on a relationship of foreign key to Table1 (column: key Table1.Primary)

  As I have no need for data, but want to keep the structure of the table, I would ideally like to delete partitions at the same time that the deletion of the data, so that the 2 tables partitioned and empty.

  Additionally, Table1 has some partitioned index I would be rebuilt as an index unpartitioned (given that Table1 will be empty).

  I thought I would start by removing all partitions of Table1 (via "ALTER TABLE DROP PARTITION nom_partition Table1"), but when the script came to the last partition, I got this error message:

  ORA-14083: cannot delete the only one partition of a partitioned table

  ORA-06512: at "SYS." DROP_PARTITIONS', line 46

  If someone could advise as to the best approach to what I want to achieve with regard to the two tables? Two tables down and recreate them then without partitions would be the easiest way?

  Thanks in advance for any guidance.

  If someone could advise as to the best approach to what I want to achieve with regard to the two tables? Two tables down and recreate them then without partitions would be the easiest way?

  Yes - delete the tables and re-create them.

  As says the exception that you cannot delete the last partition. A table is partitioned either or it is not.

  There was no interest at all to use DBMS_REDEFINITION to redefine tables because you do not want the data anyway.

• 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.