with hash index global partition

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.

• Global partitioned index hash

  Is it possible to create an index of global partitioned hash on the columns of the primary key of a table that is not partitioned itself?
  If a table has a PK constraint there an index generated automatically on the PK columns.
  Y at - it a syntax to make the overall index partitioned hash when creating the constraint of PK,.
  or syntax to MODIFY the indexes after the creation of the constraint?

  When you add a Pk to a table, you can choose a place already in the index. Something like

  alter table t1 add constraint T_PK primary key (col1) using index T_IDX;
  

  Create the partitioned index as you like with the same columns as the primary key and use the foregoing.

• Global and local index with respect to table partitioned

  Hello
  I am very confused about local and global index in a partitioned table.
  In addition, local index can be partitioned and unpartitioned.
  In addition, what is the difference between global partitioned and unpartitioned
  the index?
  
  As far as I know,
  If there is local index then there will be as many partitions as the partition table
  and local parition will index contains the index of only the lines of their corresponding scores.
  In addition, the local index maintananence is easier because drop us or create a partition, only index corresponding partition is affected then that in the case of the overall index,
  If we create a new partition of the table, to rebuild the global index.
  Am I wrong?
  
  Concerning

  Hello

  This means that when you add/drop/split/merge table any partition (maintaiance) lying below, its corresponding partition in index is automatically added/deleted/split/merged. Oracle takes care of it internally without user intervention.

  Concerning
  Anurag

• 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

• Global partitioned index shows State N/A sqldeveloper

  Hi again,
  
  I tried to create a global index divided by 4 hash partitions to be used as the primary key.
  
  I think I understand the syntax for creating an index of global hash (using 4 partitions) that is used by the primary key constraint.
  Here's my response:
  
  ALTER table vehicle_data
  Add the constraint of KEY PRIMARY C_PK_ID
  (
  ID
  )
  USING INDEX
  HASH PARTITION (LWVD_ID)
  4. THE PARTITIONS
  ENABLE;
  
  What I don't understand is this:
  When I take the sqldeveloper, go to the table and select the tab 'index' index status is indicated by n/a. When I create the index without partitioning it shows valid.
  What is the problem that I forgot?
  
  Thanks again,
  Andreas

  You forgot that it is partitioned.

  If you look in ALL_IND_PARTITIONS for the State.

• 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.

• Copy a table with its index of Oracle 12 to another instance oracle 12

  Hello

  I m using 64 bit Win8

  I have a huge table T1 (about 200 million) on user storage space and its index is on different tablespaces on DB1

  I built an other empty DB2 with the same names of storage spaces. I d like copy the table T1 of DB1 with all its indexes in DB2 so that the table will in User tablespace and the index go to their corresponding storage spaces.

  Is it possible to do?

  Currently I T1 to export into a CSV file and re - import to DB2 and build all indexes manually.

  Concerning

  Hussien Sharaf

  1. What is the exact syntax to export and import a table with its index?

  You will need to use the "Table". An export of table mode is specified by the parameter TABLES. Mode table, only a specified set of tables, partitions, and their dependent objects are unloaded. See: https://docs.oracle.com/cloud/latest/db121/SUTIL/dp_export.htm#i1007514

  2 How can I import the indexes in one tablespace other than the table itself?

  You can only export the table first without the index by specifying the EXCLUSION clause in order to exclude from the index (see: https://docs.oracle.com/cloud/latest/db121/SUTIL/dp_export.htm#i1007829) and then manually create the index and specify the different tablespace when you create.

• With parallel option swap partition

  Hi all
  
  I do a swap with parallel option partition, but his giving is not not an option valid.
  
  ALTER TABLE table_1
  P_name SWAP PARTITION
  WITH TABLE table_2
  WITHOUT VALIDATION
  PARALLEL (level 12);
  
  ERROR on line 5:
  ORA-14094: invalid ALTER TABLE EXCHANGE PARTITION option
  
  Please advice.
  
  Thank you and best regards,
  Rakesh
  
  Published by: user12003321 on March 5, 2010 17:43

  This means that we cannot run with parallel option swap partition when we do not include the index of update?

  FIX!

  cannot use PARALLEL with EXCHANGE

• Satellite l.660-12 q - would the recovery disk rebuild the new HARD disk with a new recovery partition

  Talk about confusion. Does anyone know how to get a Toshiba C660 work after the HARD drive failed and the owner has stupidly NOT followed the instructions to make a recovery disk. Will be the recovery disk (if buy you one from Toshiba) rebuild the new HARD disk with a new recovery partition and install windows 7 (original OS) and accept the product key printed on the Windows license, attached to the base of the laptop. I have read a number of suggestions, but these have been considered incorrect by others.

  Surely, Toshiba must have a way to sort this problem.

  They read these messages?

  Hey Buddy

  I don't think it's really complicated, that I could find all the information about the recovery procedure in the user's manual

  The recovery disk must be created on the first day of purchase its recommended to create one in case something would be wrong with the HARD drive.

  The recovery disk contains an image. The image is a package containing Win system, drivers, tools and all the stuff pre-installed on the notebook.
  You bought the laptop and the system has already been activated so that you have need of t the key placed at the bottom of the unit.
  In addition, the use of the recovery disc formats the drive HARD integer (partitions too) and set the laptop in the same condition as at the first day of the purchase.

  There is also another option to recover the notebook called HARD drive recovery. This HARD drive recovery requires no recovery disk. The recovery disc HARD would be to use format ONLY partition C (System).

• How to divide each unique number in a matrix by the number with an index in another matrix?

  Hello

  Basically, the calculation, I tried to do is called "dividing element by element". It is a transaction between two matrices with few passes and lines. The result matrix also has the same demension and each element is obtained by dividing the element with the same index in a matrix of the element with the index even in a different matrix. for examlpe,.

  A = [2 4 6

  3 9 6]

  B = [2 2 3

  3 3 2]

  the result will be

  C = [1, 2-2

  1 3 3]

  Is there an easy way to do another going out each element using the table to index?

  Thank you very much

  Hao Liu

  Use the 2D tables instead of the matrix data type, then just use a simple primitive division. the operation will be done by item automatically.

  

  (Generally you should always use table 2D instead of the special matrix data type. The matrix data type is most useful for linear algebra)

  (Die, he must be very careful, for example, if you multiply two 2D tables, you get a multiplication of element by element, but if you do the same operation on two matrices, LabVIEW will substitute for a real matrix multiplication, which is not the same. For the division, it seems that the matrices are divided piece-by-piece, so it might work for you directly. You should just be aware that the matrices are often treated differently. "They are a very special type of data).

• Indexing of a table with multiple indexes

  Hi all

  I just used the detector of Ridge VI on table 1 d with a threshold.  I now have an array of index I need to round to use as a real index.  My question is, with this index corresponding to the points picture, how, I take the peak values

  To give a bit of context:

  1. I have three time correlated signals.  I filter them, normalize, then add them so that I can increase the signal-to-noise ratio.

  2 pic DetectionVI gives me a table where are these pics

  3. I want my end result

  A. Signal1 [peak_indices]

  B. Signal2 [peak_indices]

  C.Signal3 [peak_indices]

  Now I think about it in the way I have d code in MATLAB which is much easier, but I would like to do this in Labview and would be very happy to any idea.

  Thank you

  -Joe

  As you said, once you have rounded tip to the nearest value locations you have an array of markings. From there on, it should be a simple matter of passing this table in a for loop that auto-index of the results that you went out to generate a table of peak values.

  

• kindly tell how to use the unique value of a table with the index 0

  kindly tell how to use the unique value of a table with the index 0

  Hi
  
   
  
  Yep, use Index Array as Gerd says. Also, using the context help ( + h) and looking through the array palette will help you get an understanding of what each VI does.
  
   
  
  This is fundamental LabVIEW stuff, perhaps you'd be better spending some time going through the basics.
  
   
  
  -CC
  

• multiplying the elements of an array with their index

  How can I multiply the elements of an array with their index values.

  Thanks in advance.

  Hi aksoy,.

  Multiply the elements with their index:

  

• Did a system restore complete with built-in recovery partition on the hard drive, the recovery went well, but the computer is still slow and freezing.

  Original title: computer always slow and freezing after clean install

  It is a compaq 2007 desktop running windows vista premium, I work on it for a friend, they said he was running slow and freezing up all the time so I just went ahead and did a system restore complete with built-in recovery partition on the hard drive, the recovery went well but still have the same problems with the computer being slow and freezing up. any help would be appreciated thanks.

  Hello

  
  Step 1: I suggest you try the steps in the Sub articles to solve the problem of vista performance.
  http://Windows.Microsoft.com/en-us/Windows-Vista/optimize-Windows-Vista-for-better-performance
  http://Windows.Microsoft.com/en-us/Windows-Vista/get-computer-speed-and-performance-information
  http://Windows.Microsoft.com/en-us/Windows-Vista/get-maximum-performance-from-Windows-Vista-from-Windows-Vista-inside-out
  Step 2: You can also check if the problem persists in the clean boot state.
  
  Put your boot system helps determine if third-party applications or startup items are causing the problem.
   
  Try the steps in step 1 in the article to put your computer in clean boot mode.
  http://support.Microsoft.com/kb/929135

  Thanks and greetings
  Umesh P - Microsoft technical support.

  Visit our Microsoft answers feedback Forum and let us know what you think.
  [If this post can help solve your problem, please click the 'Mark as answer' or 'Useful' at the top of this message.] [Marking a post as answer, or relatively useful, you help others find the answer more quickly.]