display the columns as rows in the table of non-unique key

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  Published by: 802416 on October 14, 2010 04:29
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  I really don't understand why Oracle does not apply in some cases, unique keys.

  Let's say I have a table

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  ALTER table drop constraint t_uk drop index;

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  create index idx_1 on t (t_key, id, t_type);

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

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  create index idx_1 on t (t_key, id, t_type) local;

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  Drop index idx_1;

  -5

  create an index only idx_1 on t (t_key, id, t_type);

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

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  create an index only idx_1 on t (t_key, id, t_type) local;

  ALTER table t add constraint t_uk unique (t_key) using index idx_1;

  Drop index idx_1;

  -7

  create index idx_1 on t (t_key, t_type, id) local;

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  Drop index idx_1;

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  1. first and second cases were performed with success while 4th and 5th failed.

  The only difference is that the overall index in 4th and 5th cases are unique.

  It is unclear why ' ORA-14196: specified index cannot be used to apply the constraint. ' has occurred.

  2. in the third case index is similar to that in second place with the only difference that it is local.

  Still, it is not clear why it failed with ORA-14196.

  3 and the last question why ORA-14196 arose in 8 cases, while the 7th, it didn't.

  The only difference is the local index in 8 cases is unique.

  Thank you

  Hi Alex

  (1) index 4 and 5 cannot be used to police the unique constraint as a unique index must have the same list of column under duress. Index 4, to establish in a non-unique index or to remove the column id of the index. To index 5, you will need to make a non-unique index or remove the columns id and index t_type.

  (2) 3 index cannot be used to the unique constraint of police that it is a local index and is not partitioned column (t_type) in both the definition of index/constraint. If allowed, this would require Oracle having to visit each partition to ensure that any new value of t_key is truly unique, that would not scale. Include the t_type column in the constraint and the index to both t_type and t_key as of the columns (so move the id column 3rd position) and it would be successful.

  (Index 3) 8 fails for the same reason as in 1). It's a unique index with a different column from the constraint list. Remove the id column of the index or to make it a non-unique index and it will be successful.

  In summary, a unique index MUST have the same list of column than the constraint and a local index MUST have the partitioning columns in both the definition of index/constraint.

  See you soon

  Richard Foote

  http://richardfoote.WordPress.com/

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  Hi people,
  
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  You can log in to SCOTT and try this

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• Liaison on the rows in the table based on a key element

  Hello

  I have two separate tables in my form, A and B, with the following data:

  Table A:

  COMPANY_ID	DATAITEM_1
  12345	Data1Test1
  98765	Data1Test2

  Table B:

  COMPANY_ID	DATAITEM_2
  12345	Data2Test1
  98765	Data2Test2

  In both tables, the value of the COMPANY_ID in each row is selected by the user in a drop-down list. DATAITEM_1 and DATAITEM_2 are normal textfields.

  I've linked COMPANY_ID, DATAITEM_1, and DATAITEM_2 in two tables to the matching elements in my XML schema. The form now produces data that looks like this:

  ...

  < DATA >

  < > 12345 COMPANY_ID < / COMPANY_ID >

  < DATAITEM_1 > Data1Test1 < / DATAITEM_1 >

  < / DATA >

  < DATA >

  < > 98765 COMPANY_ID < / COMPANY_ID >

  < DATAITEM_1 > Data1Test2 < / DATAITEM_1 >

  < / DATA >

  < DATA >

  < > 12345 COMPANY_ID < / COMPANY_ID >

  < DATAITEM_2 > Data2Test1 < / DATAITEM_2 >

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  < DATA >

  < > 98765 COMPANY_ID < / COMPANY_ID >

  < DATAITEM_2 > Data2Test2 < / DATAITEM_2 >

  < / DATA >

  ...

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  ...

  < DATA >

  < > 12345 COMPANY_ID < / COMPANY_ID >

  < DATAITEM_1 > Data1Test1 < / DATAITEM_1 >

  < DATAITEM_2 > Data2Test1 < / DATAITEM_2 >

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  < DATA >

  < > 98765 COMPANY_ID < / COMPANY_ID >

  < DATAITEM_1 > Data1Test2 < / DATAITEM_1 >

  < DATAITEM_2 > Data2Test2 < / DATAITEM_2 >

  < / DATA >

  ...

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  I modified my example, tables with drop-down lists can be filled in with the code as preOpen event;

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  for (var i = 0; i)< list.length;="">

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  var point = list.item (i);

  this.addItem (item.COMPANY_ID.rawValue);

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  The output event code then looks like;

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  {

  var company = xfa.resolveNode ("COMPANY. (COMPANY_ID). (RawValue_===_«_«_+_COMPANY_ID.rawValue+_»_») ");

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  }

  The new sample can be downloaded from https://acrobat.com/#d=SCOps1yz0powjzxs2L9n3w

  But you will have a lot of scripts to add the form, can they add the company duplicate lines and what happens once they have selected a company added the DATAITEM_1 (or DATAITEM_2) and then the company has changed, you need to remove the data entered or allow them to change there minds once again.

  Good luck

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  Naamas wrote:
  No,
  I already read this post!

  Then you read wrong:

  SQL> CREATE TABLE client_months
    2  (
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    4    SERV_CODE         NUMBER(4),
    5    DEBIT_CODE        NUMBER(4),
    6    PERIOD_NUM        NUMBER(2),
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   16    STAMP_DATE        DATE,
   17    STAMP_USER        VARCHAR2(15 CHAR),
   18    REGION_CODE       NUMBER(9)
   19  )
   20  /
  
  Table created.
  
  SQL> CREATE UNIQUE INDEX client_months_UK  ON client_months
    2  (SUBJ_CODE, SERV_CODE, DEBIT_CODE, PERIOD_NUM, REGION_CODE)
    3  /
  
  Index created.
  
  SQL> CREATE MATERIALIZED VIEW LOG ON client_months with rowid
    2  /
  
  Materialized view log created.
  
  SQL> CREATE MATERIALIZED VIEW client_months_mv
    2  BUILD immediate
    3  REFRESH FAST WITH ROWID ON COMMIT -- pay attention to WITH ROWID
    4  AS
    5  SELECT * FROM client_months
    6  /
  
  Materialized view created.
  
  SQL>
  

  SY.

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  https://forums.Oracle.com/thread/2151775

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