Delay in a partitioned table index population

Similar Questions

• 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))
  ;

• CTXCAT index on partitioned tables

  Hi all

  I use an oracle 11.2 database and try to use the index for a sample application oracle CTXCAT.

  I saw that oracle reference documentation "the CTXCAT index can't stand table and index partitioning, not...» "in http://docs.oracle.com/cd/B28359_01/text.111/b28303/ind.htm#BEIIEAFD

  But I was able to create a CTXCAT index on a partitioned table query to use it successfully.

  So can someone tell me exactly it meaning in the documentation of oracle developer "CTXCAT index does not support index and table partitioning?

  Thanks in advance,

  Mor.

  This simply means that you cannot create a LOCAL partitioned index on a table partitioned using the CTXCAT indextype. There is not probably the creation of a global index (ie. covering all partitions) on a partitioned table.

  The wording could be clearer.

• 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

• 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.

• Partitioning of Tables/indexes

  Hello
  
  Oracle Version 10.2.0
  O/s Version: SUSE Linux
  
  Currently, all tables and indexes are stored in the ponit of NFS mounting. I would like to know if I can partition the tables/indexes on local storage temporarily.
  
  Thank you

  Hello
  Yes you can do it. But if you have a few scores on NAS and others on the local drive, then depending on the nature of the query/DML you will see some performance issues.

  MSK

• Can I create an index on a partitioned table bitmap?

  Can I create an index on a partitioned table bitmap?

  Yes, you can. But an index on a partitioned table bitmap must be a local index.
  Image bitmap index on non partitioned tables cannot be partitioned.

• SQL, not not using index on a partitioned table

  I partitioned table that has rows over 15 million
  I have a query that selects records based on the flag = 1
  There are only 2 possible values for the flag, it is 0 or 1
  I tried two index and local bitmap index on this field.
  The query always uses full table scan
  any suggestions?

  Assuming that you have an index

  CREATE INDEX idx_maint_chg_flag
     ON a_maintenance( CASE WHEN record_changed = 1 THEN 1 ELSE NULL END);
  

  your request would be

  SELECT *
    FROM a_maintenance
   WHERE (CASE WHEN record_changed = 1 THEN 1 ELSE NULL END) = 1
  

  Of course, it may be easier to follow if you create a function is_record_changed

  CREATE FUNCTION is_record_changed( p_flag NUMBER )
    RETURN NUMBER
  IS
  BEGIN
    IF( p_flag = 1 )
    THEN
      RETURN p_flag;
    ELSE
      RETURN NULL;
    END IF;
  END;
  

  create an index on this function

  CREATE INDEX idx_maint_chg_flag
     ON a_maintenance( is_record_changed( record_changed ) );
  

  and use this function in your query

  SELECT *
    FROM a_maintenance
   WHERE is_record_changed( record_changed ) = 1
  

  Justin

• 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

• after image restore, backup partition table is different

  Hello

  He is a 2012 Macbook mid Pro with El Capitan and Bootcamp.

  After repeated with Bootcamp (Windows 10 Pro) problems I started using Image of Terabyte for Linux in order to safeguard this Macbook. Forums terabyte that people know that the Image of Linux (such as Image for Windows) is able to create a reliable backup, but also to restore - restore an entire disk image or restoring individual partitions.

  I followed their instructions precisely and activated the two options "restore first Beach" and "TPG hidden OS" (as one terabyte official, Member of the forum wrote) when restoring a partition.

  Unfortunately, it did not work! Bootcamp was no longer visible in the start screen or wasn't visible, but not start.

  What I found was that I had to keep the two options (see above) with DISABILITIES.

  Then I could restore some partition images - and 10 Windows in Bootcamp is bootable.

  Well, it was the long introduction.  I'm sorry.

  Now my question is:

  Is this something that has changed my partition table?

  Before restoring an Image for backup of Linux for the first time, I had this partition table:

  localhost: ~ root # fdisk/dev/disk0

  Disk: geometry / dev/disk0: 60801/255/63 [976773168 sectors]

  Signature: 0xAA55

  From end

  #: cyl hd s - cyl hd s id [Start - size]

  ------------------------------------------------------------------------

  1: EA 1023 254 63 - 1023 254 63 [1-409639]

  2: AF 1023 254 63 - 1023 254 63 [409640-857906448] HFS +.

  3: AB 1023 254 63 - 1023 254 63 Boot Darwin [858316088-1269536]

  * 04:07 1023 254 63-1023 254 63 [859586560-117186560] HPFS/QNX / TO THE

  BUT:

  After failed attempts to restore and work finally restore my partition table operation is different:

  Fdisk/dev/disk0

  Disk: geometry / dev/disk0: 60801/255/63 [976773168 sectors]

  Signature: 0xAA55

  From end

  #: cyl hd s - cyl hd s id [Start - size]

  ------------------------------------------------------------------------

  1: EA 0 0 2 - 1023 255 63 [1-409639]

  2: AF 25 102 15 - 1023 255 63 [409640-857906448] HFS +.

  3: AB 1023 0 57 - 1023 255 63 Boot Darwin [858316088-1269536]

  * 04:07 1023 199 8-1023 255 63 [859586560-117186560] HPFS/QNX / TO THE

  Everything seems to work fine, but I wonder if the different appearance of the partition table can be a problem?

  For example, compare:

  1: EA 1023 254 63 - 1023 254 63]

  3 o

  409639]

  TO

  1: EA 0 0 2 - 1023 255 [63 3 o 409639]

  It's the output of 'diskutil list ':

  localhost: ~ root # diskutil list

  / dev/disk0 (internal, physical):

  #: NAME SIZE TYPE IDENTIFIER

  0: GUID_partition_scheme * GB 500,1 disk0

  1: disk0s1 EFI EFI 209.7 MB

  2: Apple_HFS Macintosh HD 439,2 GB disk0s2

  3: disk0s3 Apple_Boot Recovery 650.0 MB HD

  4: Microsoft database BOOTCAMP 60.0 GB disk0s4

  And the output of 'TPG - r - vv show/dev/disk0':

  See the TPG: / dev/disk0: mediasize = 500107862016; SectorSize = 512; blocks = 976773168

  See the TPG: / dev/disk0: MBR suspicious to sector 0

  See the TPG: / dev/disk0: Pri GPT to sector 1

  See the TPG: / dev/disk0: GPT Sec at sector 976773167

  start index size summary

  0 1 MBR

  1 1 Pri GPT header

  2 32 table GPT Pri

  34 6

  40 409600 1 part TPG - C12A7328-F81F-11D2-BA4B-00A0C93EC93B

  409640 857906448 2 part TPG - 48465300-0000-11AA-AA11-00306543ECAC

  858316088 1269536 3 part TPG - 426F6F74-0000-11AA-AA11-00306543ECAC

  859585624 936

  859586560 117186560 4 part TPG - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7

  976773120 15

  976773135 32 table dry GPT

  976773167 1 dry GPT header

  Out of 'gdisk/dev/disk0:

  localhost: ~ root # gdisk/dev/disk0

  GPT fdisk (gdisk) version 1.0.1

  WARNING: Open with shared lock devices will not have their

  table partition automatically reloaded!

  Scanning partition table:

  MBR: hybrid

  BSD: absent

  APM: absent

  TPG: present

  Found a valid GPT with hybrid MBR; using GPT.

  Command (? for help): p

  Disk/dev/disk0: 976773168 sectors, GiB 465.8

  Logical sector size: 512 bytes

  Identifier (GUID) of disc: 38A2E667-6E00-4C0F-BD78-C95979E3CFAE

  Partition table contains up to 128 entries

  First usable sector is 34, last usable area is 976773134

  The partitions will be aligned with the boundaries of sector 8

  Space is free total 957 sectors (478,5 KiB)

  Starting number (sector) end (sector), Code name of size

  409639 40 1 MiB EF00 EFI 200.0

  2 409640 858316087 409.1 giB AF00 Macintosh HD

  3 858316088 859585623 619.9 MiB AB00 Recovery HD

  4 859586560 976773119 55.9 giB 0700 BOOTCAMP

  Any ideas?

  Another question:

  Is it possible to save the table to partition like that?

  DD if = / dev/sda of = gptandmbr bs = 512 count = 34

  Thank you very much

  NEUMAC

  In Fdisk, the CHS numbers are arbitrary on modern disks. They are the time of BIOS and old disks with CHS geometry. Puts BCA from Mac 0-1023 (1024 cylinders), 0-255 (256 heads) and 0-63 (64 sectors). Modern disks far exceed those numbers and would drain them. Must match what is the triplet of start/end/size and it fits in GPT and MBR. You can put arbitrary 0,0,0 for CHS and you will see all issues on GPT disks.

  Please do not use TBIL to restore W10 to circa 1970-80 disks.

• Invalid partition table appear on my HP a1700n

  Can a

  • System repair and recovery to be

  The partition table is the main hard disk index, indicating to the computer what stored where... what part of the hard disk contains the Windows operating system, which contains the files of restoration, which part contains utilities, and so forth.  If this table is damaged, the computer can not tell what is stored where.  To aid the recovery partition will not work because the computer is no longer able to see.  You would need a set of CD recovery or restoration of the manufacturer of the computer to restore it. Before you do, you must run a set of diagnostics of hard disk on the disk to make sure that this is going only to fail again.

• 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.

• partition table

  When we need to use the partition table?

  Partitioning defined

  The concept of divide and conquer has been around since the time of Sun Tzu (500 BC). Recognizing the wisdom of this concept, Oracle applied to management of large tables and indexes. Oracle has continued to evolve and refine its capabilities of partitioning since its first implementation of the Oracle 8 range partitioning. Oracle 8i and 9i, Oracle has continued to add features and new methods of partitioning. The current version of Oracle 9i Release 2 continues this tradition by adding new features to the list partitioning and partitioning method new range-list.

  When the partition

  There are two main reasons to use partitioning in an environment of VLDB (very large database). These reasons are related to improving the management and performance.

  Partitioning offers:

  • Management at the level of individual partition for data loads, creating indexes and reconstruction, and backup/restore. This may cause more quickly down, because only actively managed individual partitions are not available.
  • Performance of the queries increased by selecting only the relevant partitions. This process of weeding out the partitions that do not contain the data required by the application via a technique called the size of the partition.

  Use the partitioning:

  • When a table reaches a size "large". Large being defined according to your environment. Greater than 2 GB should always be considered for partitioning of tables.
  • What performance benefits outweigh the extra management issues related to partitioning.
  • When the archive data is on a calendar and repetitive. For example, data warehouses are usually given for a certain period of time (rolling window). Old data is then output to archive.