How to select rows with min value on a specific column

Question

I have a query:
select
  m.x1,
  round (to_date (l.y1, 'mm/dd/yyyy HH:MI:SS AM') - m.x2, 0) as numofdays
from
  table1 m,
  table2 l
where
  l.x3 = m.x3 and
  to_date (l.y1, 'mm/dd/yyyy HH:MI:SS AM') >= TO_DATE('01012013','MMDDYYYY') and
and I got this result table:
x1 (ID)
numofdays
001
5
001
10
002
2
003
3
003
1
004
0
005
66
several ID's have multiple values on the second column, I want to have only distinct IDs with smallest "numofdays" like this:
x1 (ID)
numofdays
001
5
002
2
003
1
004
0
005
66
Any ideas?

Answer 1

Hi,
The most general and versatile way is a Top-N Query:
WITH   got_r_num AS
    select
            m.x1,
            round (to_date (l.y1, 'mm/dd/yyyy HH:MI:SS AM') - m.x2, 0) as numofdays
    ,       ROW_NUMBER () OVER ( PARTITION BY  m.x1
                                 ORDER BY      to_date (l.y1, 'mm/dd/yyyy HH:MI:SS AM') - m.x2
                               )  AS r_num
    from 
          table1 m,
          table2 l
    where
          l.x3 = m.x3 and
          to_date (l.y1, 'mm/dd/yyyy HH:MI:SS AM') >= TO_DATE('01012013','MMDDYYYY') and
SELECT  x1, numofdays
FROM    got_r_num
WHERE   r_num   = 1
If you'd care to post CREATE TABLE and INSERT statements for your sample data, then I could test it.
Notice that the sub-query (got_r_num) is exactly what you posted, only with a new column (r_num) added to the SELECT clause.