Table created with partitions... user_tab_subpartitions well, but Alter table.

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• What happens to the existing after the partition of table index and created with local index

  Hi guys,.

  / / DESC part id name number, varchar2 (100), number of wage

  In an existing table PART I add 1 column DATASEQ MORE. I wonder the part of table based on dataseq.now, the table is created with this logic of partition

  create the part table partition (identification number, name varchar2 (100), number of salary, number DATASEQ) in list (dataseq) (values partition PART_INITIAL (1));

  Suggestionn necessary. given that the table is partitioned based on DATASEQ I wonder to add local indexes on dataseq. to dataseq, I have added a local index create index idx on share (dataseq) LOCAL; Now my question is, already, there are the existing index is the column ID and salary.

  (1) IDX for dataseq is created locally so that it will be partition on each partition on the main table. Please tell me what is happening to the index on the column ID and salary... it will create again in local?

  Please suggest

  S

  Hello

  first of all, in reality 'a partition table' means create a new table a migration of existing data it (although, theoretically, you can use dbms_redefinition to partition an existing table - however, it's just doing the same thing behind the scenes). This means that you also get to decide what to do with the index - index will be local, who will be global (you can also reassess some of existing indexes and decide that they are not really necessary).

  Second of all, the choice of the partitioning key seems weird. Partitioning is a data management technique more that anything else, in order to be eligible, you must find a good partitioning key. A column recently added, named "data_seq" is not a good candidate. Can you give us more details about this column and why it was chosen as a partitioning key?

  I suspect that the person who proposed this partitioning scheme made a huge mistake. A non-partitioned table is much better in all aspects (including the ease of management and performance) that divided one wrongly.

  Best regards

  Nikolai

• Need to create the Partition in the history table

  Hello

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

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

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

  Users can query for any date range.

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

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

  Please suggest and let me know for any clarification.

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

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

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

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

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

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

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

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

• Create index partition in the partition table tablespace

  Hello
  
  I am running a work custom that
  
  * Creates a tablespace by day
  * Creates the daily table partition in the created tablespace
  * Removes the days tablepartition X
  * Removes the storage space for this partition of X + 1 day.
  
  The work above works perfectly, but it has problems with the management of the index for these partitioned tables. In the old database (10g - single node), all indexes and partitions exist in a BIG tablespace and when I imported the table creation script in the new database, I changed all the partitions table & index to go in their respective space.
  
  For example:
  
  Table_name... Nom_partition... Index_Part_name... Tablespace_name
  ============...================............====================...........=================
  TABL1... TABL1_2012_07_16... TABL1_IDX_2012_07_16... TBS_2012_07_16
  TABL1... TABL1_2012_07_15... TABL1_IDX_2012_07_15... TBS_2012_07_15
  
  
  But now, when the job is run, it creates the index in the tablespace TBS_DATA default.
  
  Table_name... Nom_partition... Index_Part_name... Tablespace_name
  ============...================.............====================...........=================
  TABL1... TABL1_2012_08_16... TABL1_IDX_2012_08_16... TBS_DATA
  TABL1... TABL1_2012_08_15... TABL1_IDX_2012_08_15... TBS_DATA
  
  
  I can issue alter index rebuild to move the index to its tablespace default, but how can I make sure that the index is created in the designated tablespace?
  
  NOTE: the partition/tablespace management work that I run only creates the partition of the table and not the index.
  
  
  The new env is a cluster of CARS of 2 nodes 11 GR 2 on Linux x86_64.
  
  
  Thanks in advance,
  aBBy.

  try something like this

  ALTER table tab_owner.tab_name add the partition v_new_part_nm
  values less (to_date('''|| v_new_part_dt_formatted ||'') ((', "DD-MON-YYYY)) tablespace ' | part_tbs
  update the index (ind1_name (partition ind_partition_name tablespace ind_part_tbs)
  ind2_name (partition tablespace ind_part_tbs ind_partition_name))
  ;

• To create the partition extended, but I already have 4 partitions

  I want to create another partition to boot another OS from, but I already have 4 so I need to create an extended partition to put some of them in. How can I create an extended partition, if I already have 4?

  The four partitions that I have are: Partition OEM (39MB), Hibernation Partition (8.01 GB), OS (168 GB) Partition and the recovery Partition (12.25 GB). I have also 50GB of unallocated space.

  You cannot with an MBR disk. There is only one space in the Partition table for four partitions so no amount of tweaking can overcome that.

  You can have 4 primary or extended 3 primary partitions and 1. You have to look at the structure of your car and what you can lose.

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• How to create a partition only for my operating system another to play with me?

  My relay problem that I had towing partition C: and D: drive, but I had to remove my D: drive because my C: drive was full, and now I want to create and partition only for my O.S. and only this, so that any other program can be installed in my D: drive.

  Someone could answer my question A.S.A.P. Please, I want to know if I can do it or not, and if I can how to do it.

  Before trying to partition, which is shown in management disks for disk 0? (the name of the partitions & sizes)

  This PC came with win7 preinstalled, and if yes do you have does the recovery discs?

  Note If you attempt to partition, without first giving that info, you can trash your system and lose everything on the disc

• 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.

• 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

• 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

• I created an interactive navigation along the animated and imported and that the muse. Animations works very well, but does not have the interactivity. ((par exemple si je clique sur le "bouton à la maison", que le site ne s'ouvre pas dans mon navigateur

  I created an interactive navigation along the animated and imported and that the muse.
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