CTXCAT index on partitioned tables

Similar Questions

• creating indexes on partitioned tables...

  Hi all
  I use 11g. creates a table with partition on transaction_dt. Now I have to create index based on the combination of the column below.
  can anyone suggest the best method to create the partioned indexes based on the underside of the columns?
  
  enjoyed the great suggestions...
  

  CREATE INDEX TB_PRCHS_SALE_TRANS_BASE_IDX1 ON TB_PRCHS_SALE_TRANS_BASE
  (TRANSACTION_DT, VENDOR_ACCT_NBR, VENDOR_SECURITY_NBR);
  
  
  CREATE INDEX TB_PRCHS_SALE_TRANS_BASE_IDX2 ON TB_PRCHS_SALE_TRANS_BASE
  (TRANSACTION_DT, ING_ACCT_NBR, ING_SECURITY_NBR);
  
  
  CREATE INDEX TB_PRCHS_SALE_TRANS_BASE_IDX3 ON TB_PRCHS_SALE_TRANS_BASE
  (TRADE_DT, SETTLEMENT_DT, VENDOR_ACCT_NBR, VENDOR_SECURITY_NBR);
  
  
  CREATE UNIQUE INDEX TB_PRCHS_SALE_TRANS_BASE_PK ON TB_PRCHS_SALE_TRANS_BASE
  (TRANS_CONTROL_ID);
  
  
  CREATE INDEX TB_PRCHS_SALE_TRANS_INDX3 ON TB_PRCHS_SALE_TRANS_BASE
  (OLD_TRANSACTION_ID);
  
  
  ALTER TABLE TB_PRCHS_SALE_TRANS_BASE ADD (
    CONSTRAINT TB_PRCHS_SALE_TRANS_BASE_PK
   PRIMARY KEY
   (TRANS_CONTROL_ID));
  
  
  Table
  ====================
  
  CREATE TABLE TB_PRCHS_SALE_TRANS_BASE
  (
    TRANSACTION_DT              DATE              NOT NULL,
    TRANS_CONTROL_ID            NUMBER            NOT NULL,
    VENDOR_CD                   CHAR(3 BYTE)      NOT NULL,
    ING_ACCT_NBR                NUMBER(10)        NOT NULL,
    ING_ACCT_TYPE_CD            CHAR(1 BYTE)      NOT NULL,
    ING_SECURITY_NBR            NUMBER(10)        NOT NULL,
    VENDOR_ACCT_NBR             VARCHAR2(10 BYTE) NOT NULL,
    VENDOR_ACCT_TYPE_CD         CHAR(1 BYTE)      NOT NULL,
    VENDOR_ACCT_CHCK_CD         CHAR(1 BYTE)      NOT NULL,
    VENDOR_SECURITY_NBR         VARCHAR2(10 BYTE),
    BUY_SELL_CD                 CHAR(1 BYTE)      NOT NULL,
    TRADE_DT                    DATE,
    SETTLEMENT_DT               DATE,
    CREATE_USER                 VARCHAR2(50 BYTE),
    UPDATE_TS                   DATE,
    UPDATE_USER                 VARCHAR2(50 BYTE),
    OLD_TRANSACTION_ID          NUMBER(10)
  partition by range (TRANSACTION_DT)
  interval (numtoyminterval(1,'YEAR'))
  (
     partition p001 values less than (date '2002-01-01'),
     partition p002 values less than (date '2003-01-01'),
     partition p003 values less than (date '2004-01-01'),
     partition p004 values less than (date '2005-01-01'),
     partition p005 values less than (date '2006-01-01'),
     partition p006 values less than (date '2007-01-01'),
     partition p007 values less than (date '2008-01-01'),
     partition p008 values less than (date '2009-01-01'),
     partition p009 values less than (date '2010-01-01'),
     partition p010 values less than (date '2011-01-01'),
    );
   

  CREATE INDEX TB_PRCHS_SALE_TRANS_INDX3 ON TB_PRCHS_SALE_TRANS_BASE
  (OLD_TRANSACTION_ID)
  LOCAL
  (
  partition p001,
  partition p002,
  partition p003,
  partition p004,
  partition p005,
  partition p006,
  partition p007,
  partition p008,
  partition p009,
  partition p010
  )
  /
  -- Since PK column set does not include partitioning colums
  -- you can't create LOCAL partitioned index.
  ALTER TABLE TB_PRCHS_SALE_TRANS_BASE ADD
   CONSTRAINT TB_PRCHS_SALE_TRANS_BASE_PK
  PRIMARY KEY
  (TRANS_CONTROL_ID)
  USING INDEX
  GLOBAL
  (
  partition p001,
  partition p002,
  partition p003,
  partition p004,
  partition p005,
  partition p006,
  partition p007,
  partition p008,
  partition p009,
  partition p010
  )
  /
  

  SY.

• Partitioned global index on partitioned table range, but the index partition does not work

  Hello:

  I was creating an index partitioned on table partitioned and partitioned index does not work.

  create table table_range)

  CUST_FIRST_NAME VARCHAR2 (20).

  CUST_GENDER CHAR (1),

  CUST_CITY VARCHAR2 (30),

  COUNTRY_ISO_CODE CHAR (2),

  COUNTRY_NAME VARCHAR2 (40),

  COUNTRY_SUBREGION VARCHAR2 (30),

  PROD_ID NUMBER NOT NULL,

  CUST_ID NUMBER NOT NULL,

  TIME_ID DATE NOT NULL,

  CHANNEL_ID NUMBER NOT NULL,

  PROMO_ID NUMBER OF NON-NULL,

  QUANTITY_SOLD NUMBER (10.2) NOT NULL,

  AMOUNT_SOLD NUMBER (10.2) NOT NULL

  )

  partition by (range (time_id)

  lower partition p1 values (u01 tablespace to_date('2001/01/01','YYYY/MM/DD')),

  lower partition (to_date('2002/01/01','YYYY/MM/DD')) tablespace u02 p2 values

  );

  create index ind_table_range on table2 (prod_id)

  () global partition range (prod_id)

  values less than (100) partition p1,

  lower partition p2 values (maxvalue)

  );

  SQL > select TABLE_NAME, SUBPARTITION_COUNT, HIGH_VALUE, nom_partition NUM_ROWS of user_tab_partitions;

  TABLE_NAME NOM_PARTITION SUBPARTITION_COUNT HIGH_VALUE NUM_ROWS

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

  TABLE_RANGE P2 0 TO_DATE (' 2002-01-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIA 259418)

  TABLE_RANGE P1 0 TO_DATE (' 2001-01-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIA 659425)

  SQL > select INDEX_NAME, NUM_ROWS nom_partition, HIGH_VALUE user_ind_partitions;

  INDEX_NAME NOM_PARTITION HIGH_VALUE NUM_ROWS

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

  P1 IND_TABLE_RANGE 100 479520

  IND_TABLE_RANGE P2 MAXVALUE 439323

  SQL > EXECUTE DBMS_STATS. GATHER_TABLE_STATS (USER, 'TABLE_RANGE');

  SQL > EXECUTE DBMS_STATS. GATHER_TABLE_STATS (USER, 'TABLE_RANGE', GRANULARITY = > 'PARTITION');

  SQL > EXECUTE DBMS_STATS. GATHER_INDEX_STATS (USER, 'IND_TABLE_RANGE');

  SQL > EXECUTE DBMS_STATS. GATHER_INDEX_STATS (USER, 'IND_TABLE_RANGE', GRANULARITY = > 'PARTITION');

  SQL > set autotrace traceonly

  SQL > alter shared_pool RAS system;

  SQL > changes the system built-in buffer_cache;

  SQL > select * from table_range

  where prod_id = 127;

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

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

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

  |   0 | SELECT STATEMENT |             | 16469 |  1334K |  3579 (1) | 00:00:43 |       |       |

  |   1.  RANGE OF PARTITION ALL THE |             | 16469 |  1334K |  3579 (1) | 00:00:43 |     1.     2.

  |*  2 |   TABLE ACCESS FULL | TABLE_RANGE | 16469 |  1334K |  3579 (1) | 00:00:43 |     1.     2.

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

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

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

  2 - filter ("PROD_ID" = 127)

  Statistics

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

  320 recursive calls

  2 db block Gets

  13352 consistent gets

  11820 physical reads

  0 redo size

  855198 bytes sent via SQL * Net to client

  12135 bytes received via SQL * Net from client

  1067 SQL * Net back and forth to and from the client

  61 sorts (memory)

  0 sorts (disk)

  15984 rows processed

  Once the sentence you say ' does not ' and then you go to paste plans that seem to show that it "works".

  What gives?

  In fact, if you look at the plans - think Oracle you have 16 k rows in the table and he'll be back k 12 rows for your select statement. In this case, Oracle is picking up the right plan - full scan 16 ranks of k is a lot less work to digitize the index lines k 12 followed by the research of rank k 12 rowid.

• index of output table and show what LOCAL indexes and who are

  I read the view of ALL_INDEXES here:
  http://www.Stanford.edu/dept/ITSS/docs/Oracle/10G/server.101/b10755/statviews_1061.htm
  But I still don't know how to query follows:
  
  1. display the name of the table, the index name and flag Y/N 'Is the local index' and a flag "is the overall index.
  
  The application should output table index and show what indexes are LOCAL and which are not.
  
  How to write this query?
  
  My original request:
  

  select table_name, Index_Name, Uniqueness, Partitioned, decode(Partitioned, 'NO', 'Global','Local') GlobalOrLocal from all_indexes
  where table_name = 'MyTABLE'

  2 I understand that global index for partitioned tables is a chunk of data referencing all the partitions table in the single room, but Local index is partitioned itself there are several parts each one or more partitions to reference table?
  
  Published by: CharlesRoos on October 6, 2010 12:45 AM

  CharlesRoos wrote:
  I read the view of ALL_INDEXES here:
  http://www.Stanford.edu/dept/ITSS/docs/Oracle/10G/server.101/b10755/statviews_1061.htm
  But I still don't know how to query follows:

  1. display the name of the table, the index name and flag Y/N 'Is the local index' and a flag "is the overall index.

  The application should output table index and show what indexes are LOCAL and which are not.

  How to write this query?

  My original request:

  select table_name, Index_Name, Uniqueness, Partitioned, decode(Partitioned, 'NO', 'Global','Local') GlobalOrLocal from all_indexes
  where table_name = 'MyTABLE'
  

  That seems Ok to me.

  2 I understand that for partitioned tables index global is a big piece of data referencing all the partitions table in the single room,

  Yes.

  but the Local index is partitioned itself there several pieces for each reference one or several partitions table?

  Almost right: each refers to score exactly one table. I.e. index partitioning is 'aligned' with the partitioning of the table.

• Unusable index Partitioned table

  Oracle Version: 11.2.0.2
  OS: Linux RHEL 5
  
  I'm having a problem with the index on a partitioned table interval. This is the scenario:
  
  There are 3 indices (A, B, C) on a table. A is created on two columns (1,2), B is on the column (2), C is the column (3). Table is partitioned on column 2 apart. A and B are partitioned indexes. When I drop a partition without a clause to update index, Index C will unusable. After seeing this, I gave up the C rating and create it as a local partitioned index, then all indexes are in condition of use after the drop partition statement. Is - this mandatory on a range partitioned table all indexes on the table must be locally partitioned inorder to have indexes in usable state after the drop statement of partition (without the update index clause)? Also there will be a negative impact for all indexes on the table like locally partitioned indexes?

  >
  Is - this mandatory on a range partitioned table all indexes on the table must be locally partitioned inorder to have indexes in usable state after the drop statement of partition (without the update index clause)?
  >
  Yes - it should be obvious by looking at a simple example.

  You have an OVERALL index and drop a partition but say Oracle are NOT updating the overall index.

  How this index may be usable for anything? It is not accurate. There was GARBAGE everywhere that gets in the way of Oracle find the entries of the "good". There are the index keys in the branch blocks that belong to the partition that has been deleted. Oracle cannot use these keys to determine how the blocks of branch, he needs to find.
  >
  Also there will be a negative impact for all indexes on the table like locally partitioned indexes?
  >
  Who knows? There are maybe or maybe not.

  As with many things Oracle "it depends."

  It depends on what types of queries that you ran.

  It depends on the filter predicates used for query.

  It depends on if the maintenance simple score (add/drop/split) are important to you.

  For the maintenance of the partition only all index LOCAL is the best.

  For full table scan, an index is not intended to be used.

  For a query that needs all THE records for a COMPANY_ID given in a table partitioned by DATE an OVERALL index could be better. Oracle also use 100 index LOCAL (if there are 100 partitions) to get the same data.

  It depends on.

• 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))
  ;

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

  Can I create an index on a partitioned table bitmap?

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

• Delay in a partitioned table index population

  Hello
  
  We have a 3 TB database, which earns about 5 GB per day. Fat is partitioned by date (a partition for each day), as well as their local indexes. The population of data is made by IBM DataStage and assume that it uses a bulk loading. I have 2 questions:
  
  1. is the index updated immediately for cargo in bulk, or Oracle wait until the load has finished updating indexes on table?
  2. is there a way to tell Oracle to have an index for a table, but not for a partition of this table? What I want to do is fill partition today the table with data today, but be able to build the index for this partition at a later date. Deletion and re-creation of the index for the entire table are not an option.
  
  Thank you.

  What you would normally do here is to load the data into a separate intermediate table, build indexes on this staging of the table, and then create a swap partition to load the data into the partitioned table. Exchange partition is a way to exchange very quickly a partition of a partitioned with a separate table and unpartitioned table. This happens almost instantly.

  Justin

• SQL, not not using index on a partitioned table

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

  Assuming that you have an index

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

  your request would be

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

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

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

  create an index on this function

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

  and use this function in your query

  SELECT *
    FROM a_maintenance
   WHERE is_record_changed( record_changed ) = 1
  

  Justin

• Partitioned Tables and indexes

  Hello
  
  
  I have a question on the table and index partitioning. My scenario is:
  
  Charge 2 mio records in table T once a month. Loaded records are added to existing records, and once loaded data is never changed.
  At some point, I want to delete the older recordings, so I intend to this partition table.
  
  T table looks like:
  

  create table t (id       number(10) not null  constraint t_pk primary key,
                  period   number(10) not null,
                  contract number(10) not null,
                  attr     number(10) not null);
  
  create unique index t_ux1 on t(contract,period);
  
  create index t_ix2 on t(period);

  My plan is to partition T over the period, and I'm trying to read through the concepts
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14220/partconc.htm#g471747
  
  
  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,
  
  «1. If the table partitioning column is a subset of index keys, use a local index.»
  
  "2. If the index is unique, use a global index. If this is the case, you are finished. »
  
  
  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local
  
  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?
  
  If true, what will happen when a partion fell?
  
  
  I am new in this area, so please feel the comment as you wish.
  
  
  Concerning
  Peter
  
  
  BANNER
  ----------------------------------------------------------------
  Oracle Database 10 g Enterprise Edition release 10.2.0.3.0 - 64bi
  PL/SQL version 10.2.0.3.0 - Production
  CORE Production 10.2.0.3.0
  AMT for IBM/AIX RISC System/6000: Version 10.2.0.3.0 - production
  NLSRTL Version 10.2.0.3.0 - Production

  Peter Gjelstrup wrote:

  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,

  «1. If the table partitioning column is a subset of index keys, use a local index.»

  "2. If the index is unique, use a global index. If this is the case, you are finished. »

  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local

  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?

  A partitioned index locally can only be defined as unique if the partition key is part of the columns in the index. Imagine what the database would have to do if this is not the case: in order to verify if a newly added or updated value violates the uniqueness, it will have to travel all the partitions in a serialized operation - means that no one else could do the same thing at the same time. Since he is a killer of serious scalability in terms of locking and contention, this is not allowed.

  So: Your T_UX1 index can be defined as a unique index that is local because it contains the partition key. Although the index is not prefixed ("Prefix" means that it is divided by the left side of the columns in the index) which means that there may be access patterns where all partitions should be scanned or the optimizer cannot use a method of size of effective partition according to the way the index is reached.

  Your T_PK index cannot be set as local because it must be unique (you can not use a local non-unique index in this case), but does not contain your partition key. It must be a global index. An overall index can be partitioned as well (different from the underlying table) but it doesn't have to be.

  Depends on how you access your data you have not T_IX2 index when partitioning by this key because it corresponds to the partition key and therefore could not actually be used by the mechanism of partition pruning that limit your query to the scores of individuals.

  If you have more than one MAS environment where running queries are used longer, you should be fine with the index the in general (because they could be analyzed in parallel in parallel operations), but if you have an OLTP environment, then you should avoid local no prefix indexes due to the potential problem that you need to analyze all partitions.

  Be borne in mind that with partitioning adds an important layer of complexity to other areas: in particular the options available to the optimizer and analyze cost optimizer statistics. Depends on how you access your statistical data must be maintained on several levels now (level of score and at the global level, in the case of subpartitioning may be still at this level). If your data is important and you rely on "global" level statistics (these are always the case when the optimizer at the time analysis cannot limit access to a single partition) then in the pre - 11 g databases analyze these "global" level statistics can take a lot of time and resources, since actually , you need data several times (once for the partition and even global level).

  Presenting this partitioning may mean other potential problems in terms of execution that change (not for the better sometimes) plans and how to effectively collect statistics. Note that g 11 addresses the issue of 'statistics' by introducing the so-called "extra" global statistics. Greg Rahn wrote a [blog note | http://structureddata.org/2008/07/16/oracle-11g-incremental-global-statistics-on-partitioned-tables/] on this nice feature.

  >

  If true, what will happen when a partion fell?

  Since you're already on 10g, you can specify the database to update the scores of the local index using the UPDATE of the INDEX clause, while 9i could maintain only an overall index and it is up to you to rebuild the local index partitions after the partition DDL on the table (according to the DDL operation).

  Kind regards
  Randolf

  Oracle related blog stuff:
  http://Oracle-Randolf.blogspot.com/

  SQLTools ++ for Oracle (Open source Oracle GUI for Windows):
  http://www.sqltools-plusplus.org:7676 /.
  http://sourceforge.NET/projects/SQLT-pp/

  Published by: Randolf Geist on Sep 30, 2008 16:39

  Added statistics / optimizer warning when you use the partitioning

• after image restore, backup partition table is different

  Hello

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

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

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

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

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

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

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

  Now my question is:

  Is this something that has changed my partition table?

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

  localhost: ~ root # fdisk/dev/disk0

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

  Signature: 0xAA55

  From end

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

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

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

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

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

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

  BUT:

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

  Fdisk/dev/disk0

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

  Signature: 0xAA55

  From end

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

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

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

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

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

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

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

  For example, compare:

  1: EA 1023 254 63 - 1023 254 63]

  3 o

  409639]

  TO

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

  It's the output of 'diskutil list ':

  localhost: ~ root # diskutil list

  / dev/disk0 (internal, physical):

  #: NAME SIZE TYPE IDENTIFIER

  0: GUID_partition_scheme * GB 500,1 disk0

  1: disk0s1 EFI EFI 209.7 MB

  2: Apple_HFS Macintosh HD 439,2 GB disk0s2

  3: disk0s3 Apple_Boot Recovery 650.0 MB HD

  4: Microsoft database BOOTCAMP 60.0 GB disk0s4

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

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

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

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

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

  start index size summary

  0 1 MBR

  1 1 Pri GPT header

  2 32 table GPT Pri

  34 6

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

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

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

  859585624 936

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

  976773120 15

  976773135 32 table dry GPT

  976773167 1 dry GPT header

  Out of 'gdisk/dev/disk0:

  localhost: ~ root # gdisk/dev/disk0

  GPT fdisk (gdisk) version 1.0.1

  WARNING: Open with shared lock devices will not have their

  table partition automatically reloaded!

  Scanning partition table:

  MBR: hybrid

  BSD: absent

  APM: absent

  TPG: present

  Found a valid GPT with hybrid MBR; using GPT.

  Command (? for help): p

  Disk/dev/disk0: 976773168 sectors, GiB 465.8

  Logical sector size: 512 bytes

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

  Partition table contains up to 128 entries

  First usable sector is 34, last usable area is 976773134

  The partitions will be aligned with the boundaries of sector 8

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

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

  409639 40 1 MiB EF00 EFI 200.0

  2 409640 858316087 409.1 giB AF00 Macintosh HD

  3 858316088 859585623 619.9 MiB AB00 Recovery HD

  4 859586560 976773119 55.9 giB 0700 BOOTCAMP

  Any ideas?

  Another question:

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

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

  Thank you very much

  NEUMAC

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

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

• Invalid partition table appear on my HP a1700n

  Can a

  • System repair and recovery to be

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

• Create no partition Table partitioning

  Dear all,

  I have a table that is not partition, and it has about 20 G data... If I want that table with say DATE_M partition column

  Please can anyone suggest the best way to do it.

  Thank you

  To create a partitioned table

  (1) you must make a backup of the existing one you can take expdp or DEC

  (2) remove the table and create the same table now with partitions (a minimum score is required)

  you want a range of ex-based partition if you use a date column

  (3) loading data from the previous table into the new partitioned table, either by DEC or impdp, the inserts will be automatically in their respective partitions and you don't mention explicitly.

  Their is no other possible way to partition a non partitioned table, you can do than workarounds to minimize the downtime of the table trying different ways to load data into the partitioned table.

  You can use the enable row movement clause of create table and also use local indexes for easy maintenance partition.

• Delete the data in partitioned tables

  Hello

  Oracle 11.2.0.3.0 version, running on Linux Enterprise.

  I need to remove all the data from two tables (several 100 million lines each) that are partitioned.

  (1) table 1 is a partition table have varied to a DATE column

  (2) Table2 is a partition of reference table; partitioning is referenced on a relationship of foreign key to Table1 (column: key Table1.Primary)

  As I have no need for data, but want to keep the structure of the table, I would ideally like to delete partitions at the same time that the deletion of the data, so that the 2 tables partitioned and empty.

  Additionally, Table1 has some partitioned index I would be rebuilt as an index unpartitioned (given that Table1 will be empty).

  I thought I would start by removing all partitions of Table1 (via "ALTER TABLE DROP PARTITION nom_partition Table1"), but when the script came to the last partition, I got this error message:

  ORA-14083: cannot delete the only one partition of a partitioned table

  ORA-06512: at "SYS." DROP_PARTITIONS', line 46

  If someone could advise as to the best approach to what I want to achieve with regard to the two tables? Two tables down and recreate them then without partitions would be the easiest way?

  Thanks in advance for any guidance.

  If someone could advise as to the best approach to what I want to achieve with regard to the two tables? Two tables down and recreate them then without partitions would be the easiest way?

  Yes - delete the tables and re-create them.

  As says the exception that you cannot delete the last partition. A table is partitioned either or it is not.

  There was no interest at all to use DBMS_REDEFINITION to redefine tables because you do not want the data anyway.

• Local primary key on reference partitioned Table

  Oracle running on Red Hat Linux Rel6 11.2.0.3.

  I'm on a closed network, so the following must be typed manually.

  I have a table of documents which is essentially the following:

  (entire annual <-primary key)

  whole Source_ID,

  load_dt date,

  date of doc_dt,

  doc_info clob,

  ...)

  which is partitioned on column source_id.

  I created a partitioned table of reference as follows:

  create table doc_entities_prt)

  whole doc_entity_id

  all annual,

  whole entity_id,

  forced doc_ent_fk (annual) references to documents (annual)

  tablespace...

  allow the movement of the line

  benchmark score (doc_ent_fk);

  The annual column in the child table is not unique.  The doc_entity_id column is unique.  I want to set the primary key on doc_entities_prt as a local index on the doc_entity_id column.  I was not able to find the proper syntax to get there, and now I'm wondering if this is even possible? Any ideas appreciated.

  Oops, I missed part that you want to base the PK of this index. No, it is not possible and has nothing to do with the partitoning reference. A unique index can be partitioned only if it includes a partitioning column. You can create a non-unique without partitioning column partitioned index, but then you can't create PK supported by such an index for the same reason - oracle would not be able to verify uniquenes based on this index partition and controls of the cross-partition are not supported:

  SQL > create table documents)
  2 whole annual,
  3 whole source_id,
  load_dt date 4.
  date of doc_dt 5.
  6 doc_info clob
  7                        )
  8 partition by range (source_id)
  9      (
  10 partition p1 values less than (10),
  11 partition p2 values less than (100)
  12)
  13.

  Table created.

  SQL > create index unique documents_pk
  2 on documents (annual)
  (3) local
  partition 4 p1,
  5 partition p2
  6         )
  7.
  on documents (annual)
  *
  ERROR on line 2:
  ORA-14039: partitioning columns must be a subset of the columns of a unique key
  index

  SQL > create index documents_pk
  2 on documents (annual)
  (3) local
  partition 4 p1,
  5 partition p2
  6         )
  7.

  The index is created.

  SQL > alter table documents
  2 Add the constraint documents_pk
  3 key (annual) elementary school
  4 using index documents_pk
  5.
  change the documents table
  *
  ERROR on line 1:
  ORA-14196: specified Index cannot be used to apply the constraint.

  SQL >

  SY.