Similar Questions

• Qosmio x 500 - 14W - invalid partition table

  Hay guys,

  I wanted to start my Toshiba Quosmio x 500 - 14W PQX33E (Product Code), but the only thing I found on my desk: invalid partition table.

  In another Forum I found a threat, which means I have to restart my computer. So I tried a Reboot Cd, but found nothing. Then I tried to call the Call Center of Toshiba, but they are closed...

  For me, is the Toshiba Service miserably and I hope that some guys from the Forum can help me.

  Thank you
  Paul

  Hello

  You must install the operating system once more.
  Invalid partition table means that something is wrong with the HARD drive and the operating system cannot start.
  So if you n t have a recovery disk, you must order one:
  http://backupmedia.Toshiba.EU/landing.aspx

  PS: I made good experience with the Toshiba service comparing the service with another laptop from the manufacturer I have to say that s OK and I meet a lot of people who said that other manufacturers are not really better
  Which; s my opinion

• after image restore, backup partition table is different

  Hello

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

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

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

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

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

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

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

  Now my question is:

  Is this something that has changed my partition table?

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

  localhost: ~ root # fdisk/dev/disk0

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

  Signature: 0xAA55

  From end

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

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

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

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

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

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

  BUT:

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

  Fdisk/dev/disk0

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

  Signature: 0xAA55

  From end

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

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

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

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

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

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

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

  For example, compare:

  1: EA 1023 254 63 - 1023 254 63]

  3 o

  409639]

  TO

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

  It's the output of 'diskutil list ':

  localhost: ~ root # diskutil list

  / dev/disk0 (internal, physical):

  #: NAME SIZE TYPE IDENTIFIER

  0: GUID_partition_scheme * GB 500,1 disk0

  1: disk0s1 EFI EFI 209.7 MB

  2: Apple_HFS Macintosh HD 439,2 GB disk0s2

  3: disk0s3 Apple_Boot Recovery 650.0 MB HD

  4: Microsoft database BOOTCAMP 60.0 GB disk0s4

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

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

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

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

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

  start index size summary

  0 1 MBR

  1 1 Pri GPT header

  2 32 table GPT Pri

  34 6

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

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

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

  859585624 936

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

  976773120 15

  976773135 32 table dry GPT

  976773167 1 dry GPT header

  Out of 'gdisk/dev/disk0:

  localhost: ~ root # gdisk/dev/disk0

  GPT fdisk (gdisk) version 1.0.1

  WARNING: Open with shared lock devices will not have their

  table partition automatically reloaded!

  Scanning partition table:

  MBR: hybrid

  BSD: absent

  APM: absent

  TPG: present

  Found a valid GPT with hybrid MBR; using GPT.

  Command (? for help): p

  Disk/dev/disk0: 976773168 sectors, GiB 465.8

  Logical sector size: 512 bytes

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

  Partition table contains up to 128 entries

  First usable sector is 34, last usable area is 976773134

  The partitions will be aligned with the boundaries of sector 8

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

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

  409639 40 1 MiB EF00 EFI 200.0

  2 409640 858316087 409.1 giB AF00 Macintosh HD

  3 858316088 859585623 619.9 MiB AB00 Recovery HD

  4 859586560 976773119 55.9 giB 0700 BOOTCAMP

  Any ideas?

  Another question:

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

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

  Thank you very much

  NEUMAC

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

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

• Size of partition on interval range partitioned Table happens when SYSDATE is used in a Where Clause

  We have tables interval range partitioned on a DATE, with a partition for each day column - very standard and straight out of doc Oracle.

  A 3rd party application queries the tables to find the number of rows based on the date range that is located on the column used for the partition key.

  This application uses the date range specified from the current date - i.e. for last two days would be «...» StartDate > SYSDATE-2 "- but the partition size is irrelevant and the explain command plan shows that each partition is included."

  In presenting the request uses the date in a variable partition size location and query table is obviously much better.

  DB is 11.2.0.3 on RHEL6, and default settings - i.e. nothing that could influence the behavior of the optimizer to something unusual.

  I can't work on why this would be the case. It is very easy to reproduce with cases of simple test below.

  I would be very interested to hear any views on why it's that way and if anything can be done to allow the size of the partition to work with a query including SYSDATE because it would be difficult to get the application code has changed.

  In addition to make a case to change the code, I need an explanation of why query using SYSDATE is not advisable and I know this information.

  (1) create a simple partitioned table

  CREATETABLE part_test
     (id                      NUMBER NOT NULL,
      starttime               DATE NOT NULL,
      CONSTRAINT pk_part_test PRIMARY KEY (id)) 
  PARTITION BY RANGE (starttime) INTERVAL (NUMTODSINTERVAL(1,'day')) (PARTITION p0 VALUES LESS THAN (TO_DATE('01-01-2013','DD-MM-YYYY')));
  

  (2) fill in the rows of the table 1 million distributed among 10 partitions

  BEGIN
      FOR i IN 1..1000000
      LOOP
          INSERT INTO part_test (id, starttime) VALUES (i, SYSDATE - DBMS_RANDOM.value(low => 1, high => 10));
      END LOOP;
  END;
  /
  EXEC dbms_stats.gather_table_stats('SUPER_CONF','PART_TEST');
  

  (3) to query the Table of data from the last 2 days using SYSDATE in paragraph

  EXPLAIN PLAN FOR 
  SELECT  count(*) 
  FROM    part_test
  WHERE   starttime >= SYSDATE - 2;
  

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

  | ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

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

  |   0 | SELECT STATEMENT |           |     1.     3 ×  7895 (1) | 00:00:01 |       |       |

  |   1.  GLOBAL TRI |           |     1.     3 ×            |          |       |       |

  |   2.   RANGE OF PARTITION ITERATOR.           |   111K |   867K |  7895 (1) | 00:00:01 |   KEY | 1048575.

  |*  3 |    TABLE ACCESS FULL | PART_TEST |   111K |   867K |  7895 (1) | 00:00:01 |   KEY | 1048575.

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

  Information of predicates (identified by the operation identity card):

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

  3 - filter("STARTTIME">=SYSDATE@!-2)

  (4) now do the same query, but with SYSDATE - 2 presented as a literal value.

  This query returns the same response but very different cost.

  EXPLAIN PLAN FOR
  SELECT count(*) 
  FROM part_test
  WHERE starttime >= (to_date('23122013:0950','DDMMYYYY:HH24MI'))-2;
  

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

  | ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

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

  |   0 | SELECT STATEMENT |           |     1.     8.   131 (0) | 00:00:01 |       |       |

  |   1.  GLOBAL TRI |           |     1.     8.            |          |       |       |

  |   2.   RANGE OF PARTITION ITERATOR.           |   111K |   867K |   131 (0) | 00:00:01 |   356 . 1048575.

  |*  3 |    TABLE ACCESS FULL | PART_TEST |   111K |   867K |   131 (0) | 00:00:01 |   356 | 1048575.

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

  Information of predicates (identified by the operation identity card):

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

  3 filter ("STARTTIME" > = TO_DATE (' 2013-12-21 09:50 ',' syyyy-mm-dd hh24:mi:ss'))))

  Thank you in anticipation

  Jim

  sysdate is not constant and its value may change from time optimization of execution; but the optimizer can deduce that he will be a known running value and then produce a plan that will make the size of partition running.  This is the meaning of 'KEY' in the column pstart – the partition of departure will be known at run time and the size of partition will take place in order to eliminate the previous partitions.

  Concerning

  Jonathan Lewis

• Scans of partition take longer than a full table on a non-partitioned table scan?

  Hello

  I have a PART_TABLE range partitioned table that has 10 million documents and 10 partitions having 1 million records each. Partition is based on a column named ID, which is a sequence of 1 to 10 million.

  I created another P2_BKP table (do a select * from part_table) which has the same group of data than that of PART_TABLE, except that this table is not partitioned.

  Now, I run the same query on the tables to get a range of data. Precisely, I'm reading only the data present in 5 partitions of the partitioned tables that requires theoretically less reading than when done on non partitioned tables.

  Yet the query seems to take more time on a table that is partitioned only when running on a non-partitioned table. No reason precisely why is this the case?

  Here's the query I am trying to run on the tables and their Plans to explain corresponding.

  ASK A

  =========

  Select * from P2_BKP where id < 5000000;

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

  | ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time |

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

  |   0 | SELECT STATEMENT |        |  6573K |   720 M | 12152 (2) | 00:02:26 |

  |*  1 |  TABLE ACCESS FULL | P2_BKP |  6573K |   720 M | 12152 (2) | 00:02:26 |

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

  QUERY B

  ========

  Select * from part_table where id < 5000000;

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

  | ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

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

  |   0 | SELECT STATEMENT |            |  3983K |   436 M | 22181 (73) | 00:04:27 |       |       |

  |   1.  RANGE OF PARTITION ITERATOR.            |  3983K |   436 M | 22181 (73) | 00:04:27 |     1.     5.

  |*  2 |   TABLE ACCESS FULL | PART_TABLE |  3983K |   436 M | 22181 (73) | 00:04:27 |     1.     5.

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

  016f62a3-35cf-4f4e-B812-95774e4d662f wrote:

  Now, I run the same query on the tables to get a range of data. Precisely, I'm reading only the data present in 5 partitions of the partitioned tables that requires theoretically less reading than when done on non partitioned tables.

  Yet the query seems to take more time on a table that is partitioned only when running on a non-partitioned table. No reason precisely why is this the case?

  Have you actually run the queries, or you assume that they will take different amounts of time because the execution plans show different numbers?

  It is clear from the output of the execution plan that the statistics are not very good, since Oracle should have estimated about 5 M of lines for each request. That said, however, Oracle should have used the counties of partition block level to estimate the cost of the analysis of the 5 score - and we could expect them to a total of less space than a full copy.

  Of course, it is possible that your partitioned table has been the subject of many DML involving the line movement and line of chaining; It is possible he has been the subject of many lots of parallel or direct path loads inserts leaving much free space in each partition. It is also possible that you have created the copy in a tablespace that table of default compression!

  What are the stats for the original table partition level?   (select nom_partition, num_rows, blocks of user_tab_partitions where table_name = 'T1' partition_position order)

  Concerning

  Jonathan Lewis

• To access large partitioned tables over a database link - traps?

  Hello
  
  We are in the middle of a business acquisition, and I have a question on the use of links to database access efficiently to large tables. There are two distinct geographical database instances, both on Oracle 10.2.0.5 sitting on Linux boxes.
  
  The main forum (PSHR) contains a PeopleSoft HR system and pays and is sitting in our data center.
  
  The secondary instance (HGPAY) runs a payroll application home grown and is in a different datacenter to PSHR.
  
  The requirement is to allow PeopleSoft (PSHR) to display data of payroll (one employee at a time) targeted to the secondary instance.
  
  For example in HGPAY
  
  CREATE TABLE MY_PAY_DATA AS
  SELECT TO_CHAR (A.RN, ' 00000000') 'EMP' - it is a figure 8 leading 0 unique identifier
  '20110' | TO_CHAR (B.RN) "PAY_PRD" - it is a format of exercise more than fifteen days in the year (01-27)
  C.SOME_KEY - it is the element of remuneration being considered - effectively randomly
  , 'XXXXXXXXXXXXXXXXX' "FILLER1.
  , 'XXXXXXXXXXXXXXXXX' "FILLER2".
  , 'XXXXXXXXXXXXXXXXX' "FILLER3".
  FROM (SELECT ROWNUM 'RN' FROM DUAL CONNECT BY LEVEL < = 300) has
  , (SELECT ROWNUM 'RN' FROM DUAL CONNECT BY LEVEL < = 3) B
  (SELECT TRUNC (ABS (DBMS_RANDOM. (Random())) 'SOME_KEY' FROM DUAL CONNECT BY LEVEL < = 300) C
  ORDER OF PAY_PRD, EMP
  
  HGPAY. MY_PAY_DATA is the range partitioned on EMP (approximately 300 employees by partition) and the list below partitioned on PAY_PRD (3 pay periods a secondary partition). I limited the above create statement to represent a sub-partition of data.
  Every employee generates an average of 300 lines in this table each pay period. The table has about 180 million lines and all fifteen days more.
  
  In PSHR
  
  CREATE VIEW PS_HG_PAY_DATA (MEP, PAY_PRD SOME_KEY FILLER1, FILLER2, FILLER3)
  AS SELECT EMP, PAY_PRD SOME_KEY FILLER1, FILLER2, MY_PAY_DATA@HGPAY FILLER3
  
  PeopleSoft would then generate SQL along the lines of
  
  SELECT * FROM PS_HG_PAY_DATA WHERE EMP = '00002561' AND PAY_PRD = '201025'
  
  The link between data centers where PSHR and HGPAY sit isn't the best in the world, but I expect dozens of hits per day rather than thousands, so I think that the link must have sufficient bandwidth to meet the requirements.
  
  I tried a quick test on two instances of size production test and it works because it presents the data, when I look at the plan of the explain command that I can see that the remote database is only presenting the relevant secondary partition on PSHR rather than the entire table. Until I get in the back with a "job well done" - y at - it a witch hunt that I am absent using dblink to access the partitioned tables of big?

  Yes, it's just. A lot of it depends on exactly what happens in different "oops" scenarios - you are, for example, just burn some CPU extra until someone comes to the DBA and says "my query is slow" or saturating the network has an impact on critical applications or long random queries prevent some maintenance operations of partition.

  In my mind, the simplest possible solution (assuming you are using a username that is fixed in the database link) would be to create a profile on HGPAY for user defined for the link of database that set a value LOGICAL_READS_PER_CALL which was large enough to handle any request '' reasonable '' and low enough to quickly kill any session that has tried to do something 'stupid '. Obviously, you have to define 'stupid' in your particular environment where the scope of a 'simple reconciliation report' is not defined. If there is no political problem and you can adjust the values of profile to the wire when you encounter new reports that slowly increase what is considered '' reasonable '' is probably the most straightforward approach. If you have to put in a change request to change the parameter which must be reviewed by the Control Board change at its next quarterly meeting with the outsourced DBA seller, on the other hand, you could turn a report by 30 minutes in 30 hours over 30 days. However, in an ideal world, this is where I would start.

  Becomes more complex, you can use the resource manager to kill applications running too long on the wall clock. Since the network will almost certainly the bottleneck, it is probably unlikely that the limitation of the CPU will do much good - probably you can saturate the network with a very small amount of CPU. Limitation in my mind of the network is an additional step in complexity according to the specifics of your situation and what you are competing with.

  Justin

• Query on partitioned Table

  I have two partition
  upto_mar_2011, upto_jun_2011
  table_name: order
  1 > if I run the query as below
  Select * command where order_date = to_date (' 23/03/2011 ',' dd/mm/yyyy');
  
  Oracle will automatically search the data in the Partition upto_mar_2011
  or he fetches the data in the entire table.
  
  2 > or I give the query in the form below
  Select * enforcement of partition (upto_mar_2011) where order_date = to_date (' 23/03/2011 ',' dd/mm/yyyy');
  
  Is there a difference between this two query

  OraFighter wrote:

  Is this enough to write as below

  Select * command where order_date = to_date (' 23/03/2011 ',' dd/mm/yyyy')
  or order_date = to_date (' 23/05/2011 ',' dd/mm/yyyy');

  and CBO fetch data, only the partition where it is available.
  Wheather it is spread in a two or three partition
  It will ignore all other partitions automatically

  Fix. 'S called it the size of partition. It is the approach you use to application code. Requests should not be concerned with partitioning - should not need to know the names and other physical attributes of the table physical partition it uses.

  You can also see the size of the partition when looking at such a SQL statement execution plan.

  The only code that will focus on names of partition and the physical aspects of the table's code maintenance. For example, you maintain that a sliding window of 32 days of the date was the partitioned table. This code must determine which partitions are older than 32 days and drop them. And then to add new partitions for future treatment. Copy the following code deals with the physical layer and not the logical layer.

  It is important to keep the two apart and not to mix the physical things in the logical layer which is using the app. He's mixing means that changes to optimize performance and to take advantage of the new features which may not be at the physical layer, as it will break the application layer code.

• 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 ");

• Why not all the partition tables

  Hello
  
  In our project we propse to all partition tables (even if they are small, in MBs) in order to use the feature partiion exchange.
  But our DBAs say that maintenance will be a problem if create us partitions for each table. We said, only tablespace is fine for all partitions that we thought to maintain storage different is a problem.
  But always DBA are not approved.
  I just want to know what are the disadvantages of partitioning. Why DBA are generally against the idea of creating partitions. If we use the same storage space for all of the partitions is there any side down for creating partitions for each table?
  Is there is disadvantage?
  
  
  
  Thank you
  Pramod Garre

  Pramod Garre wrote:
  We plan to use range partitioning (in the quarters).
  There is a duty to remove a quarter of the table data, then refresh with new data as and when the user update some recored of frontend.this should arrive in real-time.

  Looks like you are having the app trigger partition management... which doesn't seem a good idea.
  It also looks like you really don't have justification for partitioning on a few tables. Far from being the "all" tables you inquired. And even that looks like your concept of the use of these scores may be wrong.

  Why not work with your databases instead of look at foreigners for the justification to fight them. Sounds to me like they are doing exactly what they were hired to do.

  We thought instead of delete and insert, if we use excahnge partitions, it will be real quick as Exchange just updated partition in the dictionary.
  On these lines, do you think that it be the disadvantages of the use of partitioning?

  Note: Delete and Insert works OK (2 minutes) compared to the swap partition (20 ms).

  And Yes, each table has PK.

  Thank you
  Pramod Garre

• Adding a new partition to a partitioned table

  Hi all
  
  I try to add a new partition to an existing table in compartmentalized but I need the new partition of 'table' to go to tablespace DATA_2011 and partition "index" to go to INDEX_2011. I tried but it wrong on what follows.
  
  CREATE TABLE employees
  (employee_id number 4 NOT NULL,
  last_name VARCHAR2 (10),
  department_id NUMBER (2))
  PARTITION OF RANGE (department_id)
  (PARTITION employees_part1 VALUES LESS THAN (11) TABLESPACE DATA_2010,)
  PARTITION employees_part2 VALUES LESS THAN (21) TABLESPACE DATA_2010;
  PARTITION employees_part3 VALUES LESS THAN (31) TABLESPACE DATA_2010);
  
  CREATE INDEX employees_local_idx ON employees (employee_id) LOCAL
  (PARTITION TABLESPACE INDEX_2010 employees_part1,
  PARTITION employees_part2 TABLESPACE INDEX_2010,
  PARTITION employees_part3 TABLESPACE INDEX_2010);
  
  
  Here's what I'm trying to run but it fails...
  ALTER table employees add PARTITION "EMPLOYEES_PART4" VALUES LESS THAN (41) DATA_2011 index tablespace tablespace INDEX_2011;
  
  ERROR on line 1:
  ORA-14020: this physical attribute may not be specified for the partition table
  
  
  
  Succeeds what follows, but in this case the index partition "EMPLOYEES_PART4" is also assigned to DATA_2011 tablespace.
  
  ALTER table employees add PARTITION "EMPLOYEES_PART4" VALUES LESS THAN (41) tablespace DATA_2011;
  
  
  Any ideas... I am currently looking in the documentation to see if it would be possible?
  
  Thanks for your time...

  Hello

  ALTER table employees add PARTITION "EMPLOYEES_PART4" VALUES LESS THAN (41) tablespace DATA_2011;

  The foregoing is the corect order and therefore succeeded.

  but in this case the index partition "EMPLOYEES_PART4" is also assigned to DATA_2011 tablespace.

  You need rebuild the index of the partition with the name desired tablespace. /

  ALTER index rebuild partition tablespace

  Anand

• importing into a partitioned table of interval 11g

  as I took export utility simple partition table 8i exp not rained so 100 k lines in there.
  
  and imported with the import utility in the interval of 11g partitioned based on the date column.
  
  There were imported, but did not what I expected...
  
  If we execute the simple insert for partition interval 11g command, it create new partition automatically according to the strategy of partition.
  
  Here's the demo...
  
  created range partitioned table on the date with shift interval column...
  
  CREATE TABLE TEST.xxx_HIST
  (
  xxx_DATE DATE NOT NULL,
  P_ROLL_CONVENTION CHAR (2),
  R_ROLL_CONVENTION CHAR (2),
  P_COMPOUNDING_IND CHAR (2),
  R_COMPOUNDING_IND CHAR (2),
  P_CALC_METHOD CHAR (2),
  R_CALC_METHOD CHAR (2),
  P_SPREAD_AMT NUMBER (28,12).
  R_SPREAD_AMT NUMBER (28,12).
  
  )
  partition by range (xxx_DATE)
  interval (numtoyminterval(3,'MONTH'))
  store (security)
  (
  values of pQ1 lower partition (to_date('2010-01-01','yyyy-mm-dd'))
  ) IN PARALLEL.
  
  
  -IMPORTED FROM ROWS IN THE TABLE...
  
  ======================================================================
  Connected to: Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - 64 bit Production
  With partitioning, OLAP, Data Mining and Real Application Testing options
  
  Export file created by EXPORT: V08.01.07 direct
  
  CAUTION: objects have been exported by SYSTEM, not by you
  
  . import of xx_ARCH in TEST objects
  . . import of 141749 lines imported from the table 'xxx_HIST '.
  Import completed successfully without warnings.
  ========================================================================
  
  
  
  -HE HAS A LOT OF DATES OF DIFF IN THERE...
  
  
  
  SQL > SELECT COUNT (DISTINCT xxx_DATE) TEST.xxx_HIST;
  
  COUNT (DISTINCT xxx_DATE)
  -----------------------------
  1371
  
  
  28-MARCH 06
  10 FEBRUARY 06
  9 FEBRUARY 05
  20 FEBRUARY 02
  3 JUNE 02
  10 MAY 04
  26 DECEMBER 03
  31 JANUARY 03
  
  xxx
  ---------
  21 JULY 08
  31 OCTOBER 05
  25 APRIL 08
  28 APRIL 08
  12 OCTOBER 06
  DECEMBER 21 07
  28 DECEMBER 04
  
  
  -BUT STILL ALL DUMPED INTO A PARTITION
  
  
  SQL > SELECT nom_partition FROM DBA_TAB_PARTITIONS WHERE TABLE_OWNER = 'TEST ';
  
  NOM_PARTITION
  ------------------------------
  PQ1
  
  It all dumped in a partition...
  
  fact partition interval 11g creates the partition automatically in function whose lines if imported... when we import lines in there...? or am I missing something?
  
  any idea guys?

  Seems to be a poor strategy for me because if I am not mistaken, there is no way to specify the order of the imported lines. If you import a line with the date max as your first row... bang, you get a range partition created for you and the rest falling.

  I think you'd be better import these data into a table in step and then by a

  insert into new_fancy_partition_table
  select *
  from old_8_temporary_imported_table
  order by date_column asc
  

  Or create the partitions manually.

  I just realized that you specify a partition in your create table statement (missed that on cursory inspection). And I think you misunderstand how the interval works... it's for values LARGER than the existing partitions ONLY...

  http://download.Oracle.com/docs/CD/E11882_01/server.112/e10592/statements_7002.htm#SQLRF01402

  "
  INTERVAL clause

  Use this clause to set the interval of partitioning the table. Range partitions are partitions based on a digital range interval or datetime. * They extend from range partitioning by commanding the database to automatically create partitions of the specified range or interval when the data inserted in the table exceed all the partitions.* range
  "

  Published by: Tubby on August 16, 2010 18:32

  Additional document link.

• Disable LOGGING for partition table based on the automatic INTERVAL

  Hello
  
  I created a database table of paritioned range interval with NOLOGGING as the default attribute for the table as well as the tablespace definition.
  
  When ORACLE automatically creates a new partition for this table, it activates the LOGGING for this automatic partition even if the table definition says anything else.
  
  How can I go about changing this behavior? Is it still possible?
  
  FYI my platform is 11 GR 1 (11.1.0.7) material on SUN SOLARIS 10 SPARC.
  
  Any help will be appreciated.

  Right, so refer to a new feature of Oracle 11 g partitioning called 'partitioning interval. "

  +"+
  + The interval partitioning: A new strategy of partitioning in Oracle Database 11g, +.
  + Interval partitioning extends the functionality of the method range to define equipartitioned +.
  + ranges using a definition of the interval. Rather than specify individual +.
  + going to explicitly, Oracle will create any partition automatically as needed.
  + every time the data of a partition are inserted for the first time. Interval +.
  + a lot of partitioning improves maneuverability of a partitioned table. For +.
  example, + a partitioned table interval could be set so that Oracle creates a +.
  + new partition for each month in a calendar year; a partition is then automatically +.
  + created for "September 2007" as soon as the first record in this month is inserted.
  + in the database. +
  + The techniques available for a partitioned table interval are interval, interval-+.
  + List, range-Hash, and interval range. +
  +"+

  In order to disable interval partitioning on the transactions table, use:
  
  ALTER TABLE transactions SET INTERVAL ();
  

  http://download.Oracle.com/docs/CD/B28359_01/server.111/b32024/part_admin.htm

  Disable partitioning interval on this table, create a procedure to run every day we'll say, to create a new partition and chop the old partition of the appropriate table. I presume that the automatic partitioning interval creates the partition with the logging by default option without checking the parameters in the table. I have seen no information on this in the Oracle documentation.

  Hope that helps.

  Ogan

• 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

• How to encrypt a USB, when I get the message "(GPT) GUID Partition Table partitioning scheme is required? '."

  I just tried to encrypt a USB drive with my Mac book (latest version) and got the message "A GUID Partition Table (GPT) partitioning schΘma required." What does this mean and suggestions on how to proceed?

  The drive must be formatted for Mac. Start in the Recvery mode. Erase the drive using disk utility (select the top icon which represents the entire disk and click clear. Partition scheme must be Mac OS (Extended) first partition GUID, which is the default value.

  If you want to install a system on it follow these steps:

  How to reinstall OS X on your Mac - Apple Support

• Satellite L630-00V - need to partition table

  Hello to you guys...

  So I really screwed up here, I did my girlfriend buy a laptop (Toshiba Satellite L630) and also a new hard drive to replace the one on the inside because they have slow (5400 RPM)

  I've used ghost to clone the drive inside the laptop to the new one I mounted in a housing that is connected via USB to the computer.

  Although I double checked everything and that the destination drive was the good the bad thing happened, the blank disc has been cloned on the full.

  I don't think that all my data are deleted, because there were at least 70 GB of commands on the computer (3 recovery partitions and Windows 7 one) and I doubt enough that it can all be erased within 5 seconds.

  I used GetBackData for NTFS witch scanned the hard drive for about 05:30 (Yes, 5 hours and a half) and well found a lot of files... If right now I copy all partitions said based on another hard drive...

  I'll have to do the scan a 2nd time to get Fat32 partitions...

  Something that would be useful for me is to have the partition table of a released factory Toshiba L630-00v (500 GB HDD) *.

  
  
  
  

  Beginning and end clusters would be really useful, I would be able to rebuild my much easier partition table

  
  And oh, no idea how to get the OEM Windows 7 cd to a new facility? They were paying for a copy of windows with their computer, their must be a way to install it...

  Thank you very much! ;)

  Hello

  > And oh, no idea how to get the OEM Windows 7 cd to a new facility? They were paying for a copy of windows with their computer, their must be a way to install it.

  Now, you can get money from Toshiba Recovery disk.

  For European models:

  https://backupmedia.Toshiba.EU/landing.aspx

  American models:

  https://www.CSD.Toshiba.com/cgi-bin/TAIS/support/JSP/serviceUnitVerification.jsp?orderType=SFF

  > my girlfriend did buy a laptop (Toshiba Satellite L630) and also a new hard drive to replace the one on the inside because they have slow (5400 RPM)

  What happens if put you the old hard drive to make a recovery disk and replace it with a new one, then... Then you will be able to restore the factory settings on the new hard drive with a recovery disk. Or you have not previously burned?