Can we compress hash partitioned table in 9.2

Hi
Can we compress the hash partitioned table? How to check the compression? Is there any way to check for the partition size after compression?
Thanks

hi
go through below link
hope it will help you.
http://www.dbazine.com/oracle/or-articles/foot6
http://www.google.ae/search?hl=en&q=compressed+hash+partition+++oracle+9i&meta=
also check in google seconed point... Table compression do and don't.
hope this helps
Taj.

Similar Messages

Need to uncompress a hash partitioned table

I am working on oracle 11.1.0.7 database on solaris.
We had issues loading data into compressed tables, so we are trying to uncompress the tables and their partitions.
When I tried uncompressing a hash partitioned table, I am getting this error
SQL> alter table sample move partition SYS_1 nocompress;
alter table sample move partition SYS_1 nocompress
ERROR at line 1:
ORA-14260: incorrect physical attribute specified for this partition
I am able to move the table, it is just the nocompress option which doesn;t work.
SQL> alter table sample move partition SYS_1;
Table altered.
Can't we uncompress a hash partitioned table which is compressed?

You have to do it in two steps
alter table sample modify partition SYS_1 nocompress;
alter table sample move partition SYS_1;Best regards
Maxim

Altering Hash Partition table

Hi ,
version: 10gR2
I have a table that built with 5 hash partitions.
Since oracle recommand that hash partition table should built with power of two
(e.g: 2,4,8,16...) partition , i would like to rebuilt the table with 8 hash partition.
Its a table that contain more then 100 milion records and its size is more then 2 giga.
I understand that there isnt any: alter table .. for this kind of issue.
Please advice me what is the best way to do this task.
Thanks.

use
ALTER TABLE <table> SPLIT PARTITION <partition> AT
(<value>) INTO ( PARTITION <partition1> TABLESPACE <tbs> , PARTITION <partition2> TABLESPACE <tbs>);to divide your partitions
bye
aldo

Design capture of hash partitioned tables

Hi,
Designer version 9.0.2.94.11
I am trying to capture from a server model where the tables are hash partitioned. But this errors because Designer only knows about range partitions. Does anyone know how I can get Designer to capture these tables and their constraints?
Thanks
Pete

Pete,
I have tried all three "current" Designer clients 6i, 9i, and 10g, at the "current" revision of the repository (I can post details if interested). I have trawled the net for instances of this too, there are many.
As stated by Sue, the Designer product model does not support this functionality (details can be found on ORACLE Metalink under [Bug No. 1484454] if you have access), if not, see excerpt below. It appears that at the moment ORACLE have no urgent plans to change this (the excerpt is dated as raised in 2001 and last updated in May 2004).
Composite partitioning and List partitioning are equally affected.
>>>>> ORACLE excerpt details STARTS >>>>>
CDS-18014 Error: Table Partition 'P1' has a null String parameter
'valueLessThan' in file ..\cddo\cddotp.cpp function
cddotp_table_partition::cddotp_table_partition and line 122
*** 03/02/01 01:16 am ***
*** 06/19/01 03:49 am *** (CHG: Pri->2)
*** 06/19/01 03:49 am ***
Publishing bug, and upping priority - user is stuck hitting this issue.
*** 09/27/01 04:23 pm *** (CHG: FixBy->9.0.2?)
*** 10/03/01 08:30 am *** (CHG: FixBy->9.1)
*** 10/03/01 08:30 am ***
This should be considered seriously when looking at ERs we should be able to
do this
*** 05/01/02 04:37 pm ***
*** 05/02/02 11:44 am ***
I have reproduced this problem in 6.5.82.2.
*** 05/02/02 11:45 am *** ESCALATION -> WAITING
*** 05/20/02 07:38 am ***
*** 05/20/02 07:38 am *** ESCALATED
*** 05/28/02 11:24 pm *** (CHG: FixBy->9.0.3)
*** 05/30/02 06:23 am ***
Hash partitioning is not modelled in repository and to do so would require a
major model change. This is not feasible at the moment but I am leaving this
open as an enhancement request because it is a much requested facility.
Although we can't implement this I think we should try to detect 'partition by
hash', output a warning message that it is not supported and then ignore it.
At least then capture can continue. If this is possible, it should be tested
and the status re-set to '15'
*** 05/30/02 06:23 am *** (CHG: FixBy->9.1)
*** 06/06/02 02:16 am *** (CHG: Sta->15)
*** 06/06/02 02:16 am RESPONSE ***
It was not possible to ignore the HASH and continue processing without a
considerable amount of work so we have not made any changes. The existing
ERROR message highlights that the problem is with the partition. To enable
the capture to continue the HASH clause must be removed from the file.
*** 06/10/02 08:32 am *** ESCALATION -> CLOSED
*** 06/10/02 09:34 am RESPONSE ***
*** 06/12/02 06:17 pm RESPONSE ***
*** 08/14/02 06:07 am *** (CHG: FixBy->10)
*** 01/16/03 10:05 am *** (CHG: Asg->NEW OWNER)
*** 02/13/03 06:02 am RESPONSE ***
*** 05/04/04 05:58 am RESPONSE ***
*** 05/04/04 07:15 am *** (CHG: Sta->97)
*** 05/04/04 07:15 am RESPONSE ***
<<<<< ORACLE excerpt details ENDS <<<<<
I (like I'm sure many of us) have an urgent immediate need for this sort of functionality, and have therefore resolved to looking at some form of post process to produce the required output.
I imagine that it will be necessary to flag the Designer meta-data content and then manipulate the generator output once it's done its "raw" generation as a RANGE partition stuff (probably by using the VALUE_LESS_THAN field as its mandatory, and meaningless for HASH partitions!).
An alternative would be to write an API level generator for this using the same flag, probably using PL/SQL.
If you have (or anyone else has) any ideas on this, then I'd be happy to share them to see what we can cobble together in the absence of an ORACLE interface to their own product.
Peter

Drop partitions in HASH partitioned table

SELECT * FROM product_component_version
NLSRTL      10.2.0.4.0     Production
Oracle Database 10g Enterprise Edition      10.2.0.4.0     64bi
PL/SQL      10.2.0.4.0     Production
TNS for Solaris:      10.2.0.4.0     ProductionI have a table which is partitioned by HASH into several partitions. I would like to remove them all the same way I can DROP partitions in a LIST or RANGE partitioned tables.
I COALESCE-d my table until it remained with only one partition. Now I've got a table with one HASH partition and I would like to remove it and to end up with unpartitioned table.
How could it be accomplished?
Thank you!

Verdi wrote:
I have a table which is partitioned by HASH into several partitions. I would like to remove them all the same way I can DROP partitions in a LIST or RANGE partitioned tables.
I COALESCE-d my table until it remained with only one partition. Now I've got a table with one HASH partition and I would like to remove it and to end up with unpartitioned table.
How could it be accomplished?
You cannot turn a partitioned table into a non-partitioned table, but you could create a replacement table (including indexes etc.) and then use the 'exchange partition' option on the partitioned table. This will modify the data dictionary so the data segments for the partition exchange names with the data segments for the new table - which gives you a simple table, holding the data, in minimum time and with (virtually) no undo and redo.
The drawback to this method is that you have to sort out all the dependencies and privileges.
Regards
Jonathan Lewis
http://jonathanlewis.wordpress.com
Author: <b><em>Oracle Core</em></b>

Modify HUGE HASH partition table to RANGE partition and HASH subpartition

I have a table with 130,000,000 rows hash partitioned as below
----RANGE PARTITION--
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY RANGE (YRMO_NBR)(
PARTITION TEST_PART_200009 VALUES LESS THAN(200009),
PARTITION TEST_PART_200010 VALUES LESS THAN(200010),
PARTITION TEST_PART_200011 VALUES LESS THAN(200011),
PARTITION TEST_PART_MAX VALUES LESS THAN(MAXVALUE)
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR, LINE_ID);
Data: -
INSERT INTO TEST_PART
VALUES ('2000',200001,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200009,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200010,'CM');
VALUES ('2006',NULL,'CM');
COMMIT;
Now, I need to keep this table from growing by deleting records that fall b/w a specific range of YRMO_NBR. I think it will be easy if I create a range partition on YRMO_NBR field and then create the current hash partition as a sub-partition.
How do I change the current partition of the table from HASH partition to RANGE partition and a sub-partition (HASH) without losing the data and existing indexes?
The table after restructuring should look like the one below
COMPOSIT PARTITION-- RANGE PARTITION & HASH SUBPARTITION --
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY RANGE (YRMO_NBR)
SUBPARTITION BY HASH (C_NBR) (
PARTITION TEST_PART_200009 VALUES LESS THAN(200009) SUBPARTITIONS 2,
PARTITION TEST_PART_200010 VALUES LESS THAN(200010) SUBPARTITIONS 2,
PARTITION TEST_PART_200011 VALUES LESS THAN(200011) SUBPARTITIONS 2,
PARTITION TEST_PART_MAX VALUES LESS THAN(MAXVALUE) SUBPARTITIONS 2
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR,LINE_ID);
Pls advice
Thanks in advance

Sorry for the confusion in the first part where I had given a RANGE PARTITION instead of HASH partition. Pls read as follows;
I have a table with 130,000,000 rows hash partitioned as below
----HASH PARTITION--
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY HASH (C_NBR)
PARTITIONS 2
STORE IN (PCRD_MBR_MR_02, PCRD_MBR_MR_01);
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR,LINE_ID);
Data: -
INSERT INTO TEST_PART
VALUES ('2000',200001,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200009,'CM');
INSERT INTO TEST_PART
VALUES ('2000',200010,'CM');
VALUES ('2006',NULL,'CM');
COMMIT;
Now, I need to keep this table from growing by deleting records that fall b/w a specific range of YRMO_NBR. I think it will be easy if I create a range partition on YRMO_NBR field and then create the current hash partition as a sub-partition.
How do I change the current partition of the table from hash partition to range partition and a sub-partition (hash) without losing the data and existing indexes?
The table after restructuring should look like the one below
COMPOSIT PARTITION-- RANGE PARTITION & HASH SUBPARTITION --
CREATE TABLE TEST_PART(
C_NBR CHAR(12),
YRMO_NBR NUMBER(6),
LINE_ID CHAR(2))
PARTITION BY RANGE (YRMO_NBR)
SUBPARTITION BY HASH (C_NBR) (
PARTITION TEST_PART_200009 VALUES LESS THAN(200009) SUBPARTITIONS 2,
PARTITION TEST_PART_200010 VALUES LESS THAN(200010) SUBPARTITIONS 2,
PARTITION TEST_PART_200011 VALUES LESS THAN(200011) SUBPARTITIONS 2,
PARTITION TEST_PART_MAX VALUES LESS THAN(MAXVALUE) SUBPARTITIONS 2
CREATE INDEX TEST_PART_IX_001 ON TEST_PART(C_NBR,LINE_ID);
Pls advice
Thanks in advance

Need help on List-Hash partition - oracle 11 feature !

Can a list-hash partitioned tabled be exchanged for a partition?
Say, the table is partitioned by list on CODE (varchar2) column and subpartitioned by a NUMBER column
i.e. create table TAB1 (ID, Code, Number)
partition by LIST (Code)
subpartition by HASH (Number)
subpartition template
( subpartition1 , subpartition2 , subpartition3)
partition part1 values ('A'),
partition part1 values ('B'),
partition part1 values ('C')
Lets say the subpartitions1,2 and 3 have values 1,2,3,4,5,6....10, how can I move only say value 1 and 2 into another table using exchange partition method? Is this possible?

>
Thanks for the reply. The db version details is as below. And I am more interested in knowing if and how can data be extracted from hash sub-partitions for a given sub-partition key value, using partition exchange. Can anyone demonstrate this or point to any article that demonstrates this? I am not even sure if something like is possible.
>
What part of my reply didn't you undertand?
Except now you are saying 'extract' where before you wanted to exchange the hash subpartition. If you exchange then the subpartition will now have NO data since it will have been exchanged with an empty table.
In a partition exchange ALL of the partition (or subpartition) is exchanged, not just part of it. So for a hash subpartition you either exchange ALL data or none of it. If you only want some of the data in the subpartition you have to query it out.
No one can provide any samples until you provide a valid sample yourself. You said your partitions have character data
partition part1 values ('A'),
partition part1 values ('B'),
partition part1 values ('C')
);But then you ask about manipulating numeric data
>
Lets say the subpartitions1,2 and 3 have values 1,2,3,4,5,6....10, how
>
Which is it?
Post the DDL for the table and show which subpartition you want to query or exchange.

Ask hash algortihm in partitioning table by hash

hi all,
I wanted to ask about the hash algortihm in partitioning table by hash .How hash algorithm evenly distribute the data on each partisi??anybody know??

A simple calculation probably isn't going to be possible. The problem is that a simple hash function is unlikely to produce the nearly uniform data distribution and a function that produces the nearly uniform data distribution is unlikely to be particularly simple to compute. ORA_HASH has to balance the desire to be quick to compute against the desire to produce nearly uniform data. I don't know what algorithm Oracle picked or where relatively they made the trade off-- I expect that the algorithm may well change across releases.
Conceptually, though, if you had a perfect hash algorithm, it would provide a perfect "fingerprint" of the data and any valid input would have an equal a priori probability of being mapped anywhere in the valid output set-- just like a perfect encryption algorithm generates encrypted output that would appear totally random. If you can look at a piece of data and know that its hash was more likely to be in one part of the output set than another, that would mean that there would be an opportunity to attack the hash-- you could probabilistically reconstruct data if you knew the hash.
If you assume that ORA_HASH is a perfect hash (it probably isn't, but it's probably close enough for this analysis so long as the range is a power of 2), then for any given input, any output is equally likely. If you create a hash partitioned table with 8 partitions, that basically equates to asking ORA_HASH to generate a hash that is between 0 and 7 and puts the data in whichever partition comes out. Since each value is equally likely to be in any of the partitions, you'd expect that the data would be equally distributed. Unless of course there is a lot of repetition in the values in the column you are partitioning by-- the hash for two identical values is identical, so that sort of repetitive data would cause the data distribution to be unequal.
To take a simplistic example, let's assume you were hashing numbers and let's say you have 8 partitions. The simplest possible hash algorithm would be "mod 8". That is, if you insert a value x, insert it into the partiition (x mod 8). If x = 2, 2 mod 8 = 2, so put it in partition 2. If x = 10, 10 mod 8 = 2, so put it in partition 2. If x = 14, 14 mod 8 = 6 so put it in partition 6. If the numeric values that you insert are uniformly distributed (not unlikely if you're dealing with large amounts of data and very likely if you're dealing with sequence-generated primary keys), all 8 partitions will be roughly equally full.
Of course, this algorithm is far from perfect-- if all the data you are inserting is a power of 2, for example, then all the odd partitions would be empty and the even numbered partitions would be full. ORA_HASH needs to be quite a bit more complex than a simple mod N in order to provide more uniform mapping even if there are patterns in the input.
Justin

Can' upgrade to non-GUID tabled partition.

Apparently I can't upgrade from Leopard to Snow Leopard because by OS disk partition doesn't use the "GUID Partition Table Scheme". However, I can't change the partition table scheme to GUID in Disk Utility, as per the install instructions, because "Volume Scheme" is grayed out on "current" and "Options" is grayed out as well.
Is there some other way to change the volume scheme, or failing that some way to upgrade my system without losing all of my current application installations and data?

Freewheeling wrote:
Thanks. What can I use to back up? I've been backing up to Time Machine, but can I restore everything from that? Shouldn't I make the drive bootable when I format it, and then restore from that... or should I just go ahead and restore from the system disk after reformatting? I kind of know the general principles but know enough to realize things may not behave exactly the way I expect.
If you trust TM, then you are backed up--assuming you have been doing full system backups. Otherwise, make a bootable clone of your internal volume on an external drive. You can use Disk Utility to do so. If you need instructions, then let me know.
The drive will be bootable after partitioning; use your Sno disk to partition (see below). And, since you are partitioning, you won't have to erase. I would not restore from the system disk. Doing so would require you to upgrade from Leo to move to Sno. After partitioning, I would prefer to install Sno cleanly and then migrate what I wanted from the clone or TM--preferably from the clone IMO. Note, however, that an upgrade install, with prior restoration from TM, is perfectly acceptable if your Leopard installation is in good condition. If it is not, then erasing and installing, with migration from a clone, will give you more options in avoiding problems with your Leopard install. For example, you could choose to migrate only your data and possibly your apps.

Creation of Hash Partitioned Global Index

Hash Partion Index creation
Hi friends,
Could you suggest me whether we can create a hash partitioned index by using syntax as below in 9i.
CREATE INDEX hgidx ON tab (c1,c2,c3) GLOBAL
PARTITION BY HASH (c1,c2)
(PARTITION p1 TABLESPACE tbs_1,
PARTITION p2 TABLESPACE tbs_2,
PARTITION p3 TABLESPACE tbs_3,
PARTITION p4 TABLESPACE tbs_4);
I am getting error ORA-14005 Missing Key word Range.
Thanks in advance for your help.

Yaseer,
Is it possible to create Non-Partitioned and Global Index on Range-Partitioned Table?
Yes
We have 4 indexes on CS_BILLING range-partitioned table, in which one is CBS_CLIENT_CODE(*local partitioned index*) and others are unknown types of index to me??
Means other 3 indexes are what type indexes ...either non-partitioned global index OR non-partitioned normal index??
You got local index and 3 non-partitioned "NORMAL" b-tree tyep indexes
Also if we create index as :(create index i_name on t_name(c_name)) By default it will create Global index. Please correct me......
Above staement will create non-partitioned index
Here is an example of creating global partitioned indexes
CREATE INDEX month_ix ON sales(sales_month)
   GLOBAL PARTITION BY RANGE(sales_month)
      (PARTITION pm1_ix VALUES LESS THAN (2)
       PARTITION pm2_ix VALUES LESS THAN (3)
       PARTITION pm3_ix VALUES LESS THAN (4)
        PARTITION pm12_ix VALUES LESS THAN (MAXVALUE));Regards

Create a partitioned table using Round Robin

Hi,
I want to create a partitioned table.
My concern is that if I use round robin now, whether I can create a primary key for this table in the future?
Thanks!

You cannot use Round Robin, if your table has primary keys.
But as you must know , now we can create a HASH Partitioning on the tables with out primary key too, i thought ii would rather test once again and comment:
Please find the screen shot below which confirms that Round Robin cannot be done on table with Primary keys.
Where as the table without primary keys am able to use Round robin. Till here it was fine for me.
Now in the second case where i created a table with Round robin and then later i added the primary key using Alter statement, then am able to have Round Robin on the table with primary keys. ( don't know what does that means )
Column screenshot:
Regards,
Krishna Tangudu

IOT or Hash partition

Hi all,
I want to insert large data into a table to be retreived later using a key column (like emp no).
To the performance point of view, which is more efficient: IOT (Index Organized Table) or Hash Partition ?

I highly appreciate your time Justin. Your explanation clarified many things to me.
However, I have small notes on your comments:
Firt:
<<IOT's tend to be useful when you have thin, tall tables (many rows, few columns) where you always want to retrieve all the rows.>>
Regarding this claim, I referred to the following sources:
1. Sybex-Oracle9i Performance Tuning book
"If you access the table using its primary key, an IOT will return the rows more quickly than a traditional table."
2. http://www.tlingua.com/articles/iot.html
For single row fetch,"IOTs could provide a substantial performance gain as well as reducing the demand for disk drives"
For Index Range Scans,"IOTs significantly outperform the standard B-tree/table model during index range scans."
3. Oracle9i Database Administratorâs Guide Release 2 (9.2)
"Index-organized tables are particularly useful when you are using applications that must retrieve data based on a primary key."
As you can see Justin, none of them mentioned the thin-tall-table fact. Did you obtain it from practical experience or from some source?
Also they all showed that IOT is most useful when retreiving based on PK.
Second:
"In general, partitioning works best when you are doing set-based processing where you can use partition elimination to concentrate on a particular subset of the data."
In Sybex-Oracle9i Performance Tuning book it is stated that "Hash partitions work best when applications retrieve the data from the partitioned table via the unique key. Range lookups on hash partitioned tables derive no benefit from the partitioning.".
I can see there is some confilict, isn't it?
Thanks in advance.

Doubt in Hash Partition

Hi friends,
I have a hash partition table with 3 parts.
My doubt is Can i use a query to see how many records are present in each parts?

SQL> CREATE TABLE my_table
a INTEGER
PARTITION BY HASH (a)
PARTITION part_1,
PARTITION part_2,
PARTITION part_3
Table created.
SQL> INSERT INTO my_table
   SELECT object_id
     FROM all_objects
43055 rows created.
SQL> SELECT COUNT (*)
FROM my_table PARTITION (part_1)
COUNT(*)
     10734
1 row selected.
SQL> SELECT COUNT (*)
FROM my_table PARTITION (part_2)
COUNT(*)
     21495
1 row selected.
SQL> SELECT COUNT (*)
FROM my_table PARTITION (part_3)
COUNT(*)
     10826

Convert Normal Table to Partition Table

Hi All,
I have a Normal Table with Millions of records and I want to convert that Normal table to Hash Partition Table.
Please let me know how to do that.
Thanks in Advance
Elan

Didn't see the exchange partition example, yes its the best practice I also agree.
If your table's data is huge better to do it with CTAS(create table as select) since this ddl can be parallel and nologging.
After creating the new partitioned table you may rename old one with its all dependencies; indexes, triggers, constraints, grants etc. So it would better to get the ddl of the heap table first with dbms_metadata.
Best regards.
Message was edited by:
TongucY

Hotswap sata drives not recognized / partition table not detected

Hello Everybody,
I'm having some trouble, recently I bought to hot swap drive bays and moved two of my hardrives into them, but I can't seem to recognize the drives unless I boot with them in the "on" position, if I turn them on later, there's no indication that they are recognized at all by the system (by this I mean that there doesn't seem to be any new sd* or hd* devices appearing in /dev)
If I reboot with the drives configure in the "on" position, then arch recognizes them, but it can't read the partition table on one of the drives; it just shows up as, e.g /dev/sda and I don't see /dev/sda1; the drive has a single partition formatted with xfs (yes I've got xfsprogs installed), and I can recognize it if I boot into Parted Magic 3.5. I wonder if it's becuase I have BSD partition table? Anway, this isn't a huge issue, becuase I'm just going to transfer the data onto an external usb drive, reformat and transfer the data back.
I really do want to figure out a solution to the hotswapping problem though, because I don't want to have reboot if I want to hotswap the drives; this completely defeats the purpose of having hot-swappable drives.
Thanks for any help you can give,
-Pseudonomous
Edit 1: Potentially useful data:
I've got an ATI SB700/SB800 sata controller according to lspci
Edit 2:
I cannot for the life of me figure out how to contact AMD for tech-support, they provide no information on these Sata controllers, I have, however, contacted my motherboard's manufacturer. If anybody knows how to get in touch with AMD/ATI support, that would also probably be helpful.
Last edited by pseudonomous (2009-09-18 00:56:28)

I think I've got it working now; thanks for everyone's help.
There was in fact some BIOS settings that I needed to change hidden under "southbridge configuration". I needed to enable AHCI and disable "SATA / IDE combined controller" or something like that.
I'm also having some trouble with my fstab setup since the hotswap sata drives are in a raid1 array, and the system doesn't like it when I boot without it present; I'll see if I can figure out a way around that; if not it probably belongs in another thread, anyway,.
Again, thanks for your help!

Can we compress hash partitioned table in 9.2

Similar Messages

Maybe you are looking for