The index of the partition.

Oracle 10.2.0.2

Hello people,

I have a partition table and index on this partition. We moved all the partitions of tablespace index A to B. We dropped A tablespace, now, when I am interested in the index structure it showing me all the partition on table space B, but the index is still pointing to A partition. And because of that during the break-up of this table, we get below error.

ORA-00959: tablespace 'A' does not exist. What I have to spend this index tablespace diff too?

Hello..

I don't think that something will be disabled because of it. You can also refer to Doc ID: Note: 408021.1

Anand

Tags: Database

Similar Questions

Size of the partition and Index Local

Hi all

Database information: 11.2.0.3 64-bit Linux. It is a 3 RAC Cluster node.

I recently partitioned a Parent and child tables as PARTITION of REFERENCE set. The parent table has been RANGE - partitioned. The first partition is on an integer column and the beaches are 0-1000, 10001-2000, 20001-30000 and so on. The secondary partition is on a date and it is partitioned for each quarter.

I created a local index as follows (faullt2vehicle, offboard_load_date, record_type, Column_Y).

Here's an example query I will carry out, where the date is covering 2 secondary partitions

select  * from gets_tool_Fault t where fault2vehicle                                      = 12195
and offboard_load_date                                 BETWEEN TO_DATE('2015-04-01 00:00:00', 'yyyy-mm-dd hh24:mi:ss')
AND TO_DATE('2015-07-01 00:00:00', 'yyyy-mm-dd hh24:mi:ss')
AND record_type                                        IN('FLT','STP','HC','EGU','EOAEQP','CEL')

When I generate the plan explain below is the output

Plan hash value: 2702285198

--------------------------------------------------------------------------------------------------------------------------
| Id  | Operation                           | Name               | Rows  | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
--------------------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                    |                    |   386 | 76428 |   244   (0)| 00:00:03 |       |       |
|   1 |  PARTITION RANGE SINGLE             |                    |   386 | 76428 |   244   (0)| 00:00:03 |     2 |     2 |
|   2 |   PARTITION RANGE ITERATOR          |                    |   386 | 76428 |   244   (0)| 00:00:03 |     7 |     8 |
|   3 |    TABLE ACCESS BY LOCAL INDEX ROWID| GETS_TOOL_FAULT    |   386 | 76428 |   244   (0)| 00:00:03 |    10 |    18 |
|*  4 |     INDEX RANGE SCAN                | GETS_TOOL_FAULT_C7 |   386 |       |     7   (0)| 00:00:01 |    10 |    18 |
--------------------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   4 - access("FAULT2VEHICLE"=12195 AND "OFFBOARD_LOAD_DATE">=TO_DATE(' 2015-04-01 00:00:00', 'syyyy-mm-dd 
              hh24:mi:ss') AND "OFFBOARD_LOAD_DATE"<=TO_DATE(' 2015-07-01 00:00:00', 'syyyy-mm-dd hh24:mi:ss'))
       filter("RECORD_TYPE"='CEL' OR "RECORD_TYPE"='EGU' OR "RECORD_TYPE"='EOAEQP' OR "RECORD_TYPE"='FLT' OR 
              "RECORD_TYPE"='HC' OR "RECORD_TYPE"='STP')

If you observe the optimizer is currently to identify the partition, and then the secondary partition, but what is confusing to me, this is why the restricted PREMISES index scan scans all secondary partitions for a given partition. The partition should be not even as a value from PARTITION RANGE ITERATOR of 7 and 8.

Interestingly, when I provide a date range which is in single partition the pruning is happening correctly

Plan hash value: 3457799067

--------------------------------------------------------------------------------------------------------------------------
| Id  | Operation                           | Name               | Rows  | Bytes | Cost (%CPU)| Time     | Pstart| Pstop |
--------------------------------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                    |                    |     1 |   197 |     4   (0)| 00:00:01 |       |       |
|   1 |  PARTITION RANGE SINGLE             |                    |     1 |   197 |     4   (0)| 00:00:01 |     2 |     2 |
|   2 |   PARTITION RANGE SINGLE            |                    |     1 |   197 |     4   (0)| 00:00:01 |     7 |     7 |
|*  3 |    TABLE ACCESS BY LOCAL INDEX ROWID| GETS_TOOL_FAULT    |     1 |   197 |     4   (0)| 00:00:01 |    16 |    16 |
|*  4 |     INDEX RANGE SCAN                | GETS_TOOL_FAULT_N1 |     1 |       |     3   (0)| 00:00:01 |    16 |    16 |
--------------------------------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   3 - filter("RECORD_TYPE"='CEL' OR "RECORD_TYPE"='EGU' OR "RECORD_TYPE"='EOAEQP' OR "RECORD_TYPE"='FLT' OR 
              "RECORD_TYPE"='HC' OR "RECORD_TYPE"='STP')
   4 - access("FAULT2VEHICLE"=12195 AND "OFFBOARD_LOAD_DATE">=TO_DATE(' 2015-04-01 00:00:00', 'syyyy-mm-dd 
              hh24:mi:ss') AND "OFFBOARD_LOAD_DATE"<=TO_DATE(' 2015-06-01 00:00:00', 'syyyy-mm-dd hh24:mi:ss'))

Thank you

Jayadeep

1. your index (faullt2vehicle, offboard_load_date, record_type, Column_Y), your query has a predicate based on a beach on the second column in this index, which means that the order of the index that must be examined is dictated by this predicate and each row in the range must be considered (filtered) to check the third predicate. Basic guideline on the column in the index you are ordering for more effective use: If you have columns where predicates are always equal put these columns at the beginning of the index, the columns where predicates are usually based on a beach go towards the end. You seem to have been misled by reviews 'the prefix local index' stick to your two columns of partitioning at the front of your index - which should NOT be an automatic choice for the local index: see this note ( https://jonathanlewis.wordpress.com/2011/02/28/prefixed/ ) comments that follow and OTN article it is linked to.

2 since you have two indexes that start the same way (this goes with the local prefix again) you made easy for the optimizer to choose the index (for you) is clearly evil. That's why behind the choice of the optimizer is probably another example of an issue that I spoke a few days ago: Re: Composite index + map selection N1 your index is probably greater than the index of C7, and the order of the entries in the C7 index probably makes the data much more dispersed that it really is (i.e. high clustering_factor).

Concerning

Jonathan Lewis

PS (update):

If there are many rows for each fault_id, then you must create an index from fault_id with 'compress 1' (or compress N if the first columns of the index N are repetitive).

When dealing with range partitions do not forget that the high value partition is never reached in the document, do so based in the form "date_col BETWEEN date1 and date2", use the form "date_col > = dateX and date_col»< datey"="" or="" you="" may="" find="" that="" you="" are="" accessing="" one="" more="" partition="" than="" you="" hope="" to,="" and="" in="" some="" cases="" that="" may="" mean="" "multiple"="" (i.e.="" 2)="" when="" you="" were="" expecting="" 1="" (and="" the="" optimizer="" is="" much="" better="" at="" estimating="" costs="" for="" a="" single="">

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

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))
;

What is the index that is created automatically on the partitioning key?

CREATE TABLE sales
( prod_id NUMBER(6)
, cust_id NUMBER
, time_id DATE
, channel_id CHAR(1)
, promo_id NUMBER(6)
, quantity_sold NUMBER(3)
, amount_sold NUMBER(10,2)
)
STORAGE (INITIAL 100K NEXT 50K) LOGGING
PARTITION BY RANGE (time_id)
( PARTITION sales_q1_2006 VALUES LESS THAN (TO_DATE('01-APR-2006','dd-MON-yyyy'))
TABLESPACE tsa STORAGE (INITIAL 20K NEXT 10K)
, PARTITION sales_q2_2006 VALUES LESS THAN (TO_DATE('01-JUL-2006','dd-MON-yyyy'))
TABLESPACE tsb
, PARTITION sales_q3_2006 VALUES LESS THAN (TO_DATE('01-OCT-2006','dd-MON-yyyy'))
TABLESPACE tsc
, PARTITION sales_q4_2006 VALUES LESS THAN (TO_DATE('01-JAN-2007','dd-MON-yyyy'))
TABLESPACE tsd
)
ENABLE ROW MOVEMENT;

N ° you must create indexes for yourself.

Index BITMAP Advisor Advisor access SQL on the Partition key
I ran SQL and SQL Access Advisor as below he recommends index bitmap on p_key... Given that the cust table is partitioned on the p_key, does make sense to create indexes on p_key?

Not sure if SQL Access Advisor look just in ' where clause conditions/predicates "and the cardinality of such columns and not look at whether the table is partitioned or not. T it?

Is it wise to create BITMAP indexes on the partition key? If so what scenario would be beneficial?
```
SELECT * FROM cust
WHERE 
    'T' = 'T' AND 
    part_key IN 
    (   2, 3, 4 )) AND 
    (   p_key, act_key) IN ( select p_key, act_key from account where act_type = 'PENDING' and p_key in (2,3,4) )
```
user4529833 wrote:
Jonathan, I have exactly one value per partition for the partition key. However, most SQLs use 'IN' as predicate of partitioning pruning so overall stats are always used and they are always a bit bland compared to the partition level stats. Then this led SQL Access advisor recommendations?

Yes, the Counselor also recommended to create Bitmap indexes on act_type... Given this does make sense to have the Bitmap index on the partition key?

I'll post the execution plan as soon as I have access to the system...

The fact that statistics are "a little stale" was not much of a difference.

The fact that Oracle has probably used statistics at the level of the tables is likely to be the underlying issue. Is the partitioned table list or you have rigged it with partitioning range: If you faked list partitioning using partitioning of the range which MAY have contributed to the issue (but it's an assumption that I have not tested).

Unofficially, the optimizer has said something like:

+ "There are 25 possible values for pkey, there are 4 possible values for act_type, there seems to be 100 different combinations - so your query will pick up X lines. 80% of these lines will be packed in Y blocks and 20% of them will be scattered through the Z blocks if we have two bitmap index, and we will have to do sequential reading of the db file SSS. If we do an analysis through the affected partitions, we do MMM db file scattered reads. What are those cheaper. » +

You know how false it's - if you know that you need not the pkey index. You also know what is the efficiency of the column act_type is in the identification of data, so you can decide whether or not an index on act_key may be useful.

Concerning
Jonathan Lewis
http://jonathanlewis.WordPress.com
http://www.jlcomp.demon.co.UK

"The premature temptation to theories of shape on the lack of data is the scourge of our profession."
Sherlock Holmes (Sir Arthur Conan Doyle) in "the Valley of fear."

Bootcamp (Win10) has not started after you changed the partition and updated at El Cap.

I have a MacBook Pro (retina, 15 inches, end of 2013), 2 GHz Intel Core i7 with 8 GB of Ram.

I had 2 system installed, Win 10 + Yosemite from the windows of beginning and the use of most of the time.

Yesterday, I ran out of disk space and passed the partition in OS X with disk utility.

While I changed the partition and I also upgraded to the El Cap.

After all, I can't start by winning 10 more, although I could see my windows in the startup disk Control Panel.

I checked some thread here and mess with command autour terminal, but nothing has worked for me.

Here is the result of diskutil:

Jerbins-MacBook-Pro: ~ the jerbinyuan list $ diskutil

/ dev/disk0 (internal, physical):

#: NAME SIZE TYPE IDENTIFIER

0: GUID_partition_scheme * GB 251,0 disk0

1: disk0s1 EFI EFI 209.7 MB

2: Apple_HFS Macintosh HD 70.1 GB disk0s2

3: disk0s3 Apple_Boot Recovery 650.0 MB HD

4: Microsoft data base BART 30.0 GB disk0s4

5: Microsoft database HOMER 150,0 GB disk0s5

Jerbins-MacBook-Pro: ~ list of cs jerbinyuan$ diskutil

No logical volume group CoreStorage found

Jerbins-MacBook-Pro: ~ jerbinyuan$ sudo TPG - vv - r show/dev/disk0

Password:

See the TPG: / dev/disk0: mediasize = 251000193024; SectorSize = 512; blocks = 490234752

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 490234751

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 136992184 2 part TPG - 48465300-0000-11AA-AA11-00306543ECAC

137401824 1269536 3 part TPG - 426F6F74-0000-11AA-AA11-00306543ECAC

768 138671360

138672128 58595328 4 part TPG - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7

197267456 292966400 5 part TPG - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7

490233856 863

490234719 32 table dry GPT

490234751 1 dry GPT header

Jerbins-MacBook-Pro: ~ jerbinyuan$ sudo fdisk/dev/disk0

Disk: geometry / dev/disk0: 30515/255/63 [490234752 sectors]

Signature: 0xAA55

From end

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

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

1: EA 1023 254 63 - 1023 254 63 [1-409639] < unknown ID >

2: AF 1023 254 63 - 1023 254 63 [409640-136992184] HFS +.

3: AB 1023 254 63 - 1023 254 63 Boot Darwin [137401824-1269536]

4: 0 b 1023 254 63-1023 254 63 [138672128-58595328] Win95 FAT-32

Jerbins-MacBook-Pro: ~ jerbinyuan$ sudo dd if = / dev/rdisk0s4 account = 1 2 >/dev/null | hexdump - C

00000000 2nd eb 58 90 42 53 44 20 20 34 34 00 02 20 20 00 | X.BSD 4.4... .|

00000010 02 00 00 00 00 00 00 20 00 f8 ff 00 00 f8 43 08 |... C. |

18 00 00000020 7th e0 03 37 00 00 00 00 00 00 02 00 00 00 |... ~... 7..........|

00000030 01 06 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |... |

00000040 80 00 29 11 14 2f 2 a 42 41 52 54 20 20 20 20 20 |..) ... / * BART |

00000050 20 20 46 41 54 33 32 20 20 20 31 c0 d0 bc 8F fa | FAT32. 1... |

00000060 00 7 c fb 8th d8 e8 00 00 83 19 bb 07 00 fc 5th c6 | | ......^.......|

00000070 c0 ac 84 74 b4 06 0e cd 10 eb 30 cd 16 cd e4 f5 |... t... 0....|

00000080 19 0D 0a 4th 6f 6F 2d 73 79 73 74 65 6 d 20 64 69 |... Nonsystem di |

00000090 73 6B 0D 0 a 50 72 65 73 73 20 61 79 20 6 6 b 65 | SK... Press any ke |

000000a 0 79 20 74 6f 20 72 65 62 6f 6f 74 0D 0 a 00 00 00 | is reboot... |

000000b0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |... |

*

000001f0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 55 aa |... U. |

00000200

Jerbins-MacBook-Pro: ~ jerbinyuan$ diskutil info/dev/disk0s5

Identifier of the instrument: disk0s5

Device node: / dev/disk0s5

Set: No.

Part of the whole: disk0

Device / media name: BOOTCAMP

Volume name: HOMER

Gendarmerie: Yes

Mount point: / Volumes/HOMER

Personality of file system: NTFS

Type (Bundle): ntfs

Name (Visible to the user): Windows NT (NTFS) file system

Partition type: Microsoft Basic Data

Operating system can be installed: No.

Media type: generic

Protocol: PCI

Smart: checked

Volume UUID: 4EC75B8E-02A8-4B26-B7F3-5EA9D0A47FA9

Disk / Partition UUID: 776C43C1-7E4C-4997-BDF5-A91FD406DEBD

Total size: 150.0 GB (149998796800 Bytes) (exactly 292966400 512-Byte-units)

Amount of free space: 18.5 GB (18479718400 Bytes) (exactly 36093200 512-Byte-units)

Device block size: 512 bytes

Allocation block size: 4096 bytes

Support read-only: No.

Volume read-only: Yes

Location of the device: internal

Removable media: no

Solid State: Yes

Thanks for help.

Please follow the steps described in [help] can not boot Windows (8.1) after partition resizing .

Windows does not start after changing the partition with external software

Hello

I messed up and followed some bad advice. Water under the bridge, but I want to recover my bootcamp partition if possible. I've seen several messages with the right process, but decided to post hear instead of assumptions. I am running OS X El Capitan on an early 2011 MacBook Pro. My 256 GB SSD is a bit pressed and tried to be too stingy with the Boot Camp partition, so my current troubles after resizing. I have 10 Windows on the bootcamp partition.

I'm going to throw a bunch of data from my survey here.

Capella: ~ list of the ronh$ diskutil

/ dev/disk0 (internal, physical):

#: NAME SIZE TYPE IDENTIFIER

0: GUID_partition_scheme * GB 250,1 disk0

1: disk0s1 EFI EFI 209.7 MB

2: Apple_CoreStorage Macintosh HD 190,5 GB disk0s2

3: disk0s3 Apple_Boot Recovery 650.0 MB HD

4: Microsoft database BOOTCAMP 58.7 GB disk0s5

/ dev/disk1 (internal, virtual):

#: NAME SIZE TYPE IDENTIFIER

0: Apple_HFS Macintosh HD GB + 190,2 disk1

Logical volume on disk0s2

B6DC811B-004D-4245-83C9-8FAA0B51FD9E

Unencrypted

Capella: ~ list of cs ronh$ diskutil

CoreStorage groups of logical volumes (1 found)

|

+ Logical Volume - Group 487E3580-D1BE-4A6D-A33C-38C8DD8ABA34

=========================================================

Name: Macintosh HD

Status: online

Dimensions: 190546800640 B (190.5 GB)

Free space: 19005440 B (19.0 MB)

|

+-< physical="" volume="">

|   ----------------------------------------------------

|   Index: 0

|   Disc: disk0s2

|   Status: online

|   Dimensions: 190546800640 B (190.5 GB)

|

+ -> 37BE876E-8E51-47C3-8D62-001EC021DE56 family of Logical Volume

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

Encryption type: no

|

+-> Logical Volume B6DC811B-004D-4245-83C9-8FAA0B51FD9E

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

Disc: disk1

Status: online

Size (Total): 190175473664 B (190.2 GB)

Reversible: Yes (no required decryption)

Name: Macintosh HD

Volume name: Macintosh HD

Content indicator: Apple_HFS

Capella: ~ ronh$ sudo TPG - vv - r show/dev/disk0

Password:

See the TPG: / dev/disk0: mediasize = 250059350016; SectorSize = 512; blocks = 488397168

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 488397167

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 372161720 2 part TPG - 53746F72-6167-11AA-AA11-00306543ECAC

372571360 1269536 3 part TPG - 426F6F74-0000-11AA-AA11-00306543ECAC

373840896 1024

373841920 114554880 5 part TPG - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7

335 488396800

488397135 32 table dry GPT

488397167 1 dry GPT header

of gdisk.

Command (? for help): p

Disk/dev/rdisk0: 488397168 sectors, GiB 232.9

Logical sector size: 512 bytes

Identifier (GUID) of disc: FD5932B4-069B-4CCE-A2EE-99AB5720CA17

Partition table contains up to 128 entries

First usable sector is 34, last usable area is 488397134

The partitions will be aligned with the boundaries of sector 8

Total free space is 1365 sectors (682,5 KiB)

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

409639 40 1 200.0 of MiB EF00 EFI system partition

2 409640 372571359 177.5 giB AF05 Macintosh HD

3 372571360 373840895 619.9 MiB AB00 Recovery HD

5 373841920 488396799 54.6 giB 0700 BOOTCAMP

It is worth spending time to recover? Thank you.

Ron

1. can you describe the problem and the steps you took to get here?

2 can scroll you down to the next MS data (after that highlighted) and press 'p' to the list of files? Press ESC, once only, to return to this list.

Not enough disk space on the Partition was ' Drive

Hello!

My PC has unique of the (160 MB) hard disk with two partitions 'C' (149 MB) and ' (6.27 GB).

The partition on this disk, which would have ' a 901MB of free space on total 6.27 GB.

The PC recovery system and partition restore points system - would be "is used for the recovery system.

So, my question relates to every time I install Norton Internet Security (NIS) 2010 or earlier versions

the free space in Partition D get size reduced by 1 or 2 or maybe 3 MB. End result is that

the PC reliability and performance shows warning that free space is less than 15%

and low disk space can cause poor performance and low virtual memory errors.

This case occurs only with the installation of Norton Internet Security (NIS).

When I install McAfee Internet Security does not reduce the free space in Partition D.

The Norton came installed on my PC system, so I prefer the NIS.

Then, why this happens only in the case of NIS and not McAfee Internet Security.

How can I stop reduction of free space on the Partition was "during the installation of Norton Internet

Security. So that warnings like Low Disk Space on Partition D do not come up.

I tried, in vain, with 'Off' system restore.

Help, please.

"Then, my question relates to every time I install Norton Internet Security (NIS) 2010 or earlier versions.

the free space in Partition D get size reduced by 1 or 2 or maybe 3 MB. End result is that

the PC reliability and performance monitor shows warning that free space is less than 15%.

Something is the addition of MBs to D: drive, and ask that this happens only when you install Norton.

Contact Norton Support:

http://www.Symantec.com/en/AU/Norton/support/index.jsp

Mick Murphy - Microsoft partner

User_sub_partitions data dictionary not updated after you rename the partition but user_tab_partitions is

Hello
Using oracle 11.2.0.3
Date range poartitioned 1 partition for each month and 4 separate secondary partitions for each month.
and executed suite
ALTER table retailer_transaction rename partition SYS_P64937 tablespace PART_201505;
ALTER table move subpartition SYS_SUBP64933 tablespace RTRN_PART_201505 retailer_transaction;
ALTER table move subpartition SYS_SUBP64934 tablespace RTRN_PART_201505 retailer_transaction;
ALTER table move subpartition SYS_SUBP64935 tablespace RTRN_PART_201505 retailer_transaction;
ALTER table move subpartition SYS_SUBP64936 tablespace RTRN_PART_201505 retailer_transaction;
ALTER table retailer_transaction change the default attributes for the partition PART201505 tablespace RTRN_PART_201505;
Check user_tab_partitions and now to see nom_partition like PART_201505 but when check user_tab_subpartitions
a still 4 secondary partitions with a score of SYS_P64937 name
Why is this and how do you ensure user_sub_partitions corresponds to user_tab_partitions?
Thank you

user5716448 wrote:

Thanks for the reply.

Œuvres by renaming the index partition as suggested.

However, one last problem to try to rebuild the local bitmap index that is marked as UNUSABLE for the affected partition is ORA-14287 cannot rebuild a partition of a composite range partitioned index.

ALTER index RTRN_SAS_IDX PART_201505 rebuild partition;

How we can rebuild the local bitmap index to ensure it can be used - would not want to delete the local bitmap index and rebuild?

Thank you

[oracle@localhost ~] $14287 oerr ora

14287, 00000, "can not REBUILD a partition of a Composite range partitioned index.

* Cause: The user attempted to rebuild a partition of a Composite range partitioned index

which is illegal

* Action: REBUILD the index partition, a subpartition at a time

Move the data from the partitions again tablespace - direct table filled by ETL

Hello
We have a large composite table date-range/hash partitioned table (using time partitioning 11.2.0.3 range)
1 partition for each month and each such partition 4 under separate partitions.
We want to move some data from most significant storage but also direct table and inserted in all the days as part of the data warehouse and large table want to ensure
operations below result in errors in the data loading procedure
Suppose also should not take long to run lines 40 million or more each month.
Select *.
Of user_tab_partitions
Where table_name = 'CUSTOMER_TRANSACTION '.
And nom_partition like 'SYS % '.
1. (1) Alter table retailer_transaction and rename partition < nom_partition > CUST_PART_ < YYYYMM >
1. (2) tablespace create dedicated for the month of the affected partition
1. ((3) alter table retailer_transaction move tablespace subpartition < sys... > < partition_tablespace > created in 2) above
ALTER table retailer_transaction change the default attributes for the partition < CUST_PART < YYYYMM > < tablespace in 2 above > tablespace)
4 global index which would eventually rebuild afterwards due to the transition if movement makes no valid.
Several local bitmap indexes that belong to partitions.
Thoughts?
Thank you

I agree in a good store of data, everything should be according to the best practices ETL or guideline. I guess it is a FACT table, because it's big enough. A method that I can share (followed in the real world big DW) is "ELP" IE Partition Exchange loading. You can find the basic architecture of the Internet.

The basic architecture is as

The source (in your case partitioned) Table--> (unpartitioned) intermediate Table--> Table possible target (partitioned)--> the cleaning activities (and overtime).

You can use any ETL tool for this (I won't mention any other reasons).

Later, you can delete the original table and use that new table as your FACT table, your application will be available all the time. There are drawbacks as most uses of double-spaced, other than normal ETL these work will take certain amount of resources so that other ALS etc. In addition, if you load a few current days during these activities you plan / program accordingly.

Your demo is also fine, but regular ETL may fail because you rename the partitions (some uses of ETL tool partition name, in case you ever ELP) also unusable global indexes can cause failure (you have local too?).

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

Unique key on the partitioned table range

Hello

We use a range of composite range-hash partitioned table

Use index - try to do this have same tablespace to the local partitions but not liking it

ALTER table RETAILER_TRANSACTION_COMP_POR
Add primary key constraint RETAILER_TRANSACTION_COMP_PK (DWH_NUM)
using index
LOCAL

ORA-14039: partitioning columns must be a subset of the columns in a unique index key

Without local then fine but does not have same tablespace to walls and don't want to make this part of the partition key.

Range Tbal partitioned - it's just a UK to avoid duplicates

[oracle@localhost ~]$ oerr ora 14039
14039, 00000, "partitioning columns must form a subset of key columns of a UNIQUE index"
// *Cause:  User attempted to create a UNIQUE partitioned index whose
//          partitioning columns do not form a subset of its key columns
//          which is illegal
// *Action: If the user, indeed, desired to create an index whose
//          partitioning columns do not form a subset of its key columns,
//          it must be created as non-UNIQUE; otherwise, correct the
//          list of key and/or partitioning columns to ensure that the index'
//          partitioning columns form a subset of its key columns

increase the partition from NT 4.0

Hello
We have ESX 3.5 and there is no place free on partition of the NT 4.0 SP6a Server, installed as a guest VM.
HOWTO increase the partition from NT 4.0 without data damage?

In this case, it may be worth your while to use VMware Coverter. However, you can try to use a helper VM, but...

http://vmprofessional.com/index.php?content=2k3resize

Level lock table while truncating the partition?
Below, I use the version of oracle.

Oracle Database 11 g Enterprise Edition Release 11.2.0.2.0 - 64 bit Production
PL/SQL Release 11.2.0.2.0 - Production
CORE Production 11.2.0.2.0
AMT for Linux: Version 11.2.0.2.0 - Production
NLSRTL Version 11.2.0.2.0 - Production

I have a script to truncate the partition as below. Is there a table-level lock while truncating the parition? Any input is appreicated.

ALTER TABLE TEMP_RESPONSE_TIME TRUNCATE PARTITION part1
>
Is there a table-level lock while truncating the parition?
>
No - it will lock the partition being truncated.

Is there a global index on the table? If so they will be marked UNUSABLE and must be rebuilt.

See the VLDB and partitioning Guide
http://Oracle.Su/docs/11g/server.112/e10837/part_oltp.htm
>
Impact of surgery Maintenance of Partition on a Table partitioned local index
Whenever a partition maintenance operation takes place, Oracle locks the partitions of the table concerned for any DML operation. Data in the affected partitions, except a FALL or a TRUNCATION operation, are always fully available for any operation selection. Given that clues them are logically coupled with the (data) table partitions, only the local index partitions affected table partitions must be kept as part of a partition maintenance operation, which allows the optimal treatment for the index maintenance.

For example, when you move an old score a level of high-end storage at a level of low-cost storage, data and index are always available for SELECT operations. the maintenance of the necessary index is either update the existing index partition to reflect the new physical location of the data or, more commonly, relocation and reconstruction of the index to a level of storage partition low cost as well. If you delete an older partition once you have collected it, then its local index partitions deleted, allowing a fraction of second partition maintenance operation that affects only the data dictionary.

Impact of surgery Maintenance of Partition on Global Indexes

Whenever a global index is defined on a table partitioned or not partitioned, there is no correlation between a separate partition of table and index. Therefore, any partition maintenance operation affects all global indices or index partitions. As for the tables containing indexes, the affected partitions are locked to prevent the DML operations against the scores of the affected table. However, unlike the index for the local index maintenance, no matter what overall index remains fully available for DML operations and does not affect the launch of the OLTP system. On the conceptual and technical level, the index maintenance for the overall index for a partition maintenance operation is comparable to the index maintenance which would become necessary for a semantically identical DML operation.

For example, it is semantically equivalent to the removal of documents from the old partition using the SQL DELETE statement to drop old partition. In both cases, all the deleted data set index entries must be removed to any global index as a maintenance operation of normal index that does not affect the availability of an index to SELECT and DML operations. In this scenario, a drop operation represents the optimal approach: data is deleted without the expense of a conventional DELETE operation and the indices are maintained in a non-intrusive way.

The index of the partition.

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