Suppression of zeros of a string value

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  I have a table with ID (14 digit string values) starting with "A". Ex: A21849B1020792. There may be a different ID with the same substring 'B1020792' in the same table, example: A12349B1020792. If this happens, the ID of the last creation date of must be returned in the result. In other words, comparison is done on the same table. "In the example provided, say A12349B1020792 has the date of 1 January 2015 'and A21849B1020792 has date February 1, 2015", since A21849B1020792 has the most recent date, the result must contain only A21849B1020792.

  ID create_date

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

  A21849B1020792 02/01/2015

  A12349 B1020792 01/01/2015

  A12345B1234567 01/03/2015      

  A43567B1234567 01/04/2015      

  
  Here's the query I used:

  substr (a.id, 7) select sc, max (a.create_date) cd1

  of table a, table b

  where a.id! = b.id

  and substr (a.id, 7) = substr (b.id, 7)

  and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

  and a.id ('A12349B1020792', 'A21849B1020792','A12345B1234567 ','A43567B1234567')

  Group of substr (a.id, 7);

  Result:

  ---------

  sc                    cd1

  ---                    ------

  B1020792 02/01/2015

  B1234567 04/01/2015

  There is one thing that I'm not able to do with my request, that is, be able to get real IDs list and not list of substrings. If I added a.id column in the select clause and the group by clause, the query includes both by substr (a.id, 7) and a.id and as a result, the query returns four lines as below:

  substr (a.id, 7) select sc, max (a.create_date) cd1, a.id ID

  of table a, table b

  where a.id! = b.id

  and substr (a.id, 7) = substr (b.id, 7)

  and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

  and a.id ('A12349B1020792', 'A21849B1020792','A12345B1234567 ','A43567B1234567')

  Group by substr (a.id, 7), a.id.

  Result:

  ---------

  sc                    cd1                      ID

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

  01/01/2015 A12349B1020792 B1020792

  B1020792 01/02/2015 A21849B1020792

  B1234567 01/03/2015 A12345B1234567

  B1234567 01/04/2015 A43567B1234567

  
  

  My goal is to retrieve a list of all the unique identifiers that have the later dates.

  
  

  ID:

  ---

  A21849B1020792

  A43567B1234567

  
  

  Thanks for your time in advance!
  

  Hello

  user11951344 wrote:

  I have a table with ID (14 digit string values) starting with "A". Ex: A21849B1020792. There may be a different ID with the same substring "B1020792" in the same table, example: A12349B1020792. If this is the case, the ID of the last creation date must be returned in the result. In other words, comparison is done on the same table. "In the example provided, say A12349B1020792 updated 1 January 2015 ' and A21849B1020792 a date February 1, 2015", as A21849B1020792 has the most recent date, the result should contain only the A21849B1020792.

  ID create_date

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

  A21849B1020792 02/01/2015

  A12349B1020792 01/01/2015

  A12345B1234567 03/01/2015

  A43567B1234567 01/04/2015

  Here's the query I used:

  substr (a.id, 7) select sc, max (a.create_date) cd1

  of table a, table b

  where a.id! = b.id

  and substr (a.id, 7) = substr (b.id, 7)

  and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

  and a.id ('A21849B1020792', 'A12345B1234567', 'A12349B1020792', 'A43567B1234567')

  Group of substr (a.id, 7);

  Result:

  ---------

  sc                    cd1

  ---                    ------

  B1020792 02/01/2015

  B1234567 01/04/2015

  There is one thing that I am not able to do with my request, that is, the ability to retrieve the list of IDs real and not a list of substrings. If I added a.id column in the select clause and the group by clause, the query groups according to the two substr (a.id, 7) and a.id and as a result, the query returns four lines as below:

  substr (a.id, 7) select sc, max (a.create_date) cd1, a.id ID

  of table a, table b

  where a.id! = b.id

  and substr (a.id, 7) = substr (b.id, 7)

  and a.id like 'A %' and b.id like 'A %' / * (I used a % because I am interested only IDs that begin with A) * /.

  and a.id ('A21849B1020792', 'A12345B1234567', 'A12349B1020792', 'A43567B1234567')

  Group by substr (a.id, 7), a.id.

  Result:

  ---------

  sc                    cd1                      ID

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

  B1020792 01/01/2015 A12349B1020792

  B1020792 02/01/2015 A21849B1020792

  B1234567 03/01/2015 A12345B1234567

  B1234567 01/04/2015 A43567B1234567

  My goal is to retrieve a list of all the unique identifiers that have the later dates.

  ID:

  ---

  A21849B1020792

  A43567B1234567

  Thanks for your time in advance!

  If it makes sense to treat the first 6 characters of the ID separately from the rest of the id in this problem, maybe it makes sense to store those two parts of the id in two different columns.  Relational databases work best when each column of each row contains 1 single piece of data (at most).  It is so fundamental to the design of table that he called the first normal form.

  Given that the two parts are stored in column 1, so you can something like this Request Top - N:

  WITH got_r_num AS

  (

  SELECT r.id

  r.create_date AS cd1

  , RANK () OVER (PARTITION OF SUBSTR (r.id, 7))

  ORDER BY r.create_date DESC

  ) AS r_num

  FROM table_x g - g for data values

  JOIN table_x r - r for related values

  ON SUBSTR (r.id, 7) = SUBSTR (g.id, 7)

  WHERE g.id IN ('A12349B1020792'

  , "A21849B1020792".

  , "A12345B1234567".

  , "A43567B1234567".

  )

  - AND g.id LIKE 'A %' - if necessary.   The above condition ensures already g.id start with "A".

  AND r.id LIKE 'a % '.

  )

  SELECT id, create_date

  OF got_r_num

  WHERE r_num = 1

  ;

  If you would care to post CREATE TABLE and INSERT instructions for the sample data, and then I could test this.

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