Partitioning table and Tablespaces

Similar Questions

• How to find the partitioned tables and the number of sheets in each table

  Hello friends,
  
  I have a scheme called ICS_OWNER where I partitioned and tables not partitoined.
  
  I want to list all the tables that are partitioned.
  
  and also
  
  I want to know what partitions exists in every partitioned table.
  
  with respective schema specified
  
  Thank you/Kumar

  When you have questions like this, you don't ask here.
  but you take your (or should I say 'your') keyboard
  and type
  Select *.
  dict
  where table_name like ' % PART %.
  /

  All the dictionary views are listed in the DICT.

  ----------
  Sybrand Bakker
  Senior Oracle DBA

• Update of the partition table and update rank with php.

  Hello

  I have a partition table on my site which is

  Scores: id - userid - ranking - points-

  Get points for doing things while using the site users. Points are updated, but I don't know how

  to update this user's position by comparing everyone elses points and rank numbers update

  based on the users with most points in php.

  Hope you can help. Thank you

  Weird, the link I provided is not what I wanted; It seems very different and is difficult to read. Anyway, you don't mention what goes wrong with the code you provided. I also don't think that you want to prepend rank alias with the table name 'notes' because it is not a table column. I do not use MySQL, but I think you want to use ' @' for your SQL local variables, so something like:

  $sqlgo = mysql_query ("SET @rank = 0; SELECT @rank: = @ row + 1 rank Users.id, Users.username, Users.profilepic, Scores.scoreid, Scores.userid, Scores.level, Scores.points, rank, Scores.badges OF Scores INNER JOIN users ON Scores.userid = Users.id ORDER BY invoking DESC LIMIT 30 ");

• 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

• Question about full scan of the Table and Tablespaces

  Good evening (or morning).
  
  I read the Concepts of Oracle (I am new to Oracle) and it seems that, based on the way in which Oracle allocates and manages the storage of the local the following is true:
  
  Principle: A table which is often accessible by using a full table scan (for some reason any) would be better resident in its own dedicated tablespace.
  
  The main reason I came to this conclusion is that when you perform a full table scan, Oracle doesn't diluvium IO, reading probably one step at a time. If datafile (s) of storage space only contain data for a single table, then a reading series will have to skip segments containing the data in the other tables (as is the case if the tablespace is shared with other tables). The performance improvement is probably low, but it seems that there is a very similarly.
  
  I wish I had the thoughts of DBA experienced about the assumption above.
  
  Thank you for your contribution,
  
  John.

  a reading series will not jump the segments that contain data from other tables

  You are referring to a misleading picture of the operation.

  Let's say 'A' table is a table among a number of tables in a tablespace that consists of one or more data files.
  Since this is one of the many tables, it may not start at the beginning of a data file. His first measure may be somewhere in the middle of the file. The next step may not be nearby with the first step. The 3rd degree can be in different data file.
  So, it's the picture where you see Oracle having to jump the other tables when you do a FullTableScan.

  Now, view the same image:
  Measure 1 is in the 1000 to the 1127 block block (IE 128 blocks) in the data file 6
  Measure 2 is the 2400 block 2527 block in the data file 6
  3 is to the block to block 9371 9245 in datafile 12

  How Oracle does not have a FullTableScan? It starts with the segment header to get a map of the extent (it would read UET$ in a dictionary managed tablespace). He knows now what degrees it should read.
  What is doing?

  He made a call to the operating system to read 1000 blocks Oracle 1127 in the data file 6.
  The operating system then translates this call to starts and ID ranges and the block of the file system.
  The operating system puts this request to the storage subsystem.
  The storage subsystem and then translated the BONE, ID, blockrange to the drive starts, track, sector.
  Assume that these are all the neighbor on the single disc (that is, the volume is not distributed over multiple LUNS and disks).
  Storage reads the information and sends it to the operating system. This can still mean multiple readings, because the head (or several heads because the disc is made up of several trays) does not read 1 MB in an "appeal".
  The operating system then collects the 1 MB and passes it to Oracle.
  Oracle then fills buffers in its cache buffers--(PS: Astuce!) Tip! (: don't you know that it will be not be neighbor memory locations?)
  Microseconds past milliseconds.
  The disc is still spinning (it stops ). It has "evolved" - a different set of blocks are now under the read head.

  Oracle is now asking the OS for blocks of 2400 to 2527 in the data file 6.
  Oracle did not to "jump the other tables. He just made a separate call to the OS with a new range of blocks.
  The operating system now translates the block ID and filenumber at its own mapping.
  Storage then made his translation.
  The disc under the head blocks are very (almost certain) to be a different from blocks together. Not even those corresponding to 1128 Oracle Oracle datafile block 6. Not even the ones corresponding to 2400 block Oracle in Oracle data files.
  We are suffering now fetch - and reading the disk.

  and the story continues...

  And what happens if there is finally another user on the same system? Between the first blocks 1000-1127 call and the second call to blocks 2400-2527, that other user requested blocks 2000 (IE made a call from reading one piece).
  And, after the second measure is read by the user FullTableScan, a third user has requested block 2048 in a single block read call.

  It is important that the Table is not in the neighbor extensions? Even if you did put the table in "adjacent areas" they are not likely to be contiguous on the disk. Even if they are nearby on the disk, they are not likely to be able to read without delay, because the disc has passed.
  And we have not even spoken interleaving where readings from multiple disks need to be coordinated.

  So why still prefer 1Mo diluvium readings in Oracle? Because most SOAs allow a single OS call of up to 1 MB. Therefore, the overhead of Oracle making a call to read 1 MB are experienced only once. If the operating system allowed a maximum of 256KB reads diluvium, even a reading of 1 MB measurement must be carried out as 4 separate calls to the OS (and, recursively, 4 separate calls to the OS for the storage subsystem). We try to reduce the number of calls from software that we do. We cannot reduce the time of access and reading (but allows us to achieve more high "parallelism" in readings striping).

  Hemant K Collette
  http://hemantoracledba.blogspot.com

• 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)

• 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

• 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

• Recovery of a partition table lost on a VMFS-5 with GPT volume?

  Hello

  I have 5 ESXi

  due to a will (partition), probably an iSCSI LUN that is still visible under the target cannot be seen as data store not more

  Help all data, servers and virtual machines are lost... (they are probably still, but not seen more)   How to recover or to repeat again in the data store?

  PSE HELP

  I saw this article http://kb.vmware.com/selfservice/microsites/search.do?cmd=displayKC & docType = kc & externalId = 1002281 & sliceId = 2 & docTypeID = DT_KB_1_1 & dialogID = 291642884 & stateId = 1 0 292764722

  manages VMFS3 and the use of Fdisk u

  Can someone tell how to do it with Partutil?

  TNX for quick response

  [email protected]

  Why don't you ask before if you are unsure about the procedure?

  Run the two commands after that other was a bad idea

  Here's what I do now:

  create a volume dd image and store it somewhere safe

  1 delete the partition again - once again using partedUtil
  2 check the partition table and make sure it is clean
  3. recreate the partition with the command B
  4. start system with a USB of UNUSED ESXi or even better with a LiveCD ESXi5 to rule out any missmatches signatures
  5. try turning up the volume manually, if it is not detected automatically

  If this does not work

  1 delete the partition again - once again using partedUtil
  2 check the partition table and make sure it is clean
  3. recreate the partition with the command
  4. start system with a USB of UNUSED ESXi or even better with a LiveCD ESXi5 to rule out any missmatches signatures
  5. try turning up the volume manually, if it is not detected automatically

• partion tables and indexes

  Hi all
  
  DB: 10.2.0.4
  
  Is that we can convert a normal table into partitioned tables?
  If Yes, then what is the existing indexes on this table?
  
  IM totally new about partitioning (Table and Index). I read the oracle docs but still have confusion with local and global index (score and no-partion)...
  
  
  Please suggest...
  
  Kind regards.

  >
  Is that we can convert a normal table into partitioned tables?
  If Yes, then what is the existing indexes on this table?
  >
  You can use the EXCHANGE PARTITION to do. See this article from the Oracle base
  Partitioning of an existing Table using the EXCHANGE PARTITION
  http://www.Oracle-base.com/articles/Misc/partitioning-an-existing-table-using-Exchange-partition.php

  Or you can use for DBMS_REDEFINITION do directly

  See this article from the Oracle base
  http://www.Oracle-base.com/articles/Misc/partitioning-an-existing-table.php

  You can also create a new table and INSERT the data from the old table.

  The option you are using could depend on indexes how you have, if you intend to keep them all and if the indexes on the partitioned table must be either local or global.

  To refine the new partitioned table, you really need evaluate each index to determine that the index should be global or local, and if the index should also be partitioned.

  There isn't any point to create a new table with the same indexes if you want to redefine all the indexes. You would be better of the backup of the original and then table drop indexes before conversion.

  See the VLDB and partitioning Guide
  http://docs.Oracle.com/CD/B28359_01/server.111/b32024/TOC.htm

• Select for a partitioned table, it will work with > sign?

  Hello
  I partitioned table and select for that:

  ...
  PARTITION BY RANGE  ( "INDATE"  )
    (   PARTITION "PRT_20091201" VALUES LESS THAN (TO_DATE(' 2009-12-02 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN')...
  ....
     SELECT COLUMN_LIST FROM TABLE1   WHERE
             ( indate >= to_date(t_indate || t_timefrom, 'mm/dd/yyyy HH24:MI:SS'))                       AND
             ( indate <= to_date(t_indate || t_timeto,   'mm/dd/yyyy HH24:MI:SS'))                        AND... 

  so by selecting all of the Info for the given date specified inside time, and I use > = condition on the column partitioned, do you think that it will work correctly learn directly to the single partition or Oralce requires me to put strictly = sign for this column and then additioanly I can probably choose time, something like:
  

  SELECT COLUMN_LIST FROM TABLE1   WHERE
            ( indate = to_date(t_indate , 'mm/dd/yyyy'))                        AND
                 ( trim_to_time(indate) >= to_char(t_timefrom, ' HH24:MI:SS'))              AND
                 ( trim_to_time(indate) <= to_char( t_timeto,   ' HH24:MI:SS'))              AND... 

  All TX
  TR

  Trento wrote:
  Thanks to all the guys,
  Always learn to follow the plan. I can't use SELECT FROM TABLE of PARTITION (P - NAME) because of dynamic political anti on our site.

  Not a bad policy as dynamic sql is a blade against double edge that can cut you, and the problem with its employees.

  But do not use the option of PARTITION in a dynamic query - use manually in a one-time ad hoc when query data. Decide what partition to get the data is done automatically and is what is the process for

• Oracle Excessive Partition table

  People,
  We have a server Oracle 10 g. My question concerns an Oracle table that until now has 183 scores. Whoever created this table, put in place to add a new partition at the end of each month. I checked and verified in each partition, there are anywhere between 20 000 and 40 000 lines. Each line consists of 218 bytes. My manager is to get rid of these partitions. He just wants me to check that let down the barriers would not impact the response time that much. I can see Adding a partition of all ranks from 500 000 to 1 000 000, but having a partition for 20, 000 to 40 000 lines seems excessive to me. At the pace partitions are added to this table, in 5 years, it might have more than 600 scores. Finally, like anything else in life, there must be some back of draw for Oracle table partitioning. I saw a lot of 'Pros' on the Web but not 'jerks '.
  
  I would appreciate your opinion on it.
  
  Seyed

  I agree with your idea that the partitioning for 40K lines (with the information provided) seems excessive. Do the calculation lines 40 K * 218 partitions should be out only about 8 million, I would not immediately consider a candidate for partitioning.

  Partitioning of the benefits that you are aware of; most of them involves the use of large data sets. Pruning is a huge advantage by excluding the vast amounts of data that are unwanted review of the join.

  There are disadvantages too, for the most part buried under the volume of data partitions work usually with. There is a performance slightly beat on standard queries against a partitioned table while SQL decides which partitions to hit. You can observe this by querying the table, then by a specific partition trying only to get data for a specific partition.

  There is also the question of how many partitions you want to keep track of? You ask the right questions.

  Its possible that the partitions help performance; the only way to be sure is to create a table that is not partitioned using DEC, indexes, and constraints such as the partitioned table and run performance tests.

• Move the partition of the table and get ORA-14006: invalid partition name

  I'm using oracle 11.2.0.4 and I am trying to move a partitioned table to one tablespace to another.  I checked may times and I have the correct table name and the name of the partition.  However, I get the error ora-14006.

  Can see everything what could be the problem?

  SQL > ALTER TABLE GWPROD. QRY_TES_ROLLINGCUREDITS MOVE PARTITION 201112 TABLESPACE GW_PROD_T2 PARALLEL (DEGREE 4) NOLOGGING;

  ALTER TABLE GWPROD. QRY_TES_ROLLINGCUREDITS MOVE PARTITION TABLESPACE GW_PROD_T2 PARALLEL (DEGREE 4) NOLOGGING 201112

  *

  ERROR on line 1:

  ORA-14006: invalid partition name

  Thanks in advance.

  Names that begin with numbers are not legal partition names. A fool created by placing them between double quotes. You will need to do the same.

• Possible to move data in a partitioned into another tablespace table without indexes becoming UNUSABLE?

  Hello

  We use oracle 11.2.0.3 and know that if move us data into partitions to a different tablespace indexes get marked as UNUSABLE.

  It is the range of composite date hash partitioned table.

  As a result, to rebuild the global and local bitmap index.

  Is it possible to transfer data to one another without the index tablespace becomes UNUSABLE?

  Thank you

  Hello

  If you want to update the index (so that it becomes invalid) while you perform the move partition, you can use the index update clause.

  Details are mentioned in the Oracle documentation. Glue one of the page here for quick reference.

  Maintenance of Partitions

  Note: It has an impact on the duration of this sql statement which may increase or decrease depending on the amount of data to move.

  Concerning

  Anurag

• 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))
  ;