sql-server - Twice Inner Join on same table with Aggregate function

Question

being a novice sql user:

I have a simple table storing some records over night daily. table:

Table: T1

+----+-----+----+-----------+------------+
| Id |  A  | AB |   Value   |    Date    |
+----+-----+----+-----------+------------+
|  1 | abc | I  | -48936.08 | 2013-06-24 |
|  2 | def | A  | 431266.19 | 2013-06-24 |
|  3 | xyz | I  | -13523.90 | 2013-06-24 |
|  4 | abc | A  | 13523.90  | 2013-06-23 |
|  5 | xyz | I  | -13523.90 | 2013-06-23 |
|  6 | def | A  | 13523.90  | 2013-06-22 |
|  7 | def | I  | -13523.90 | 2013-06-22 |
+----+-----+----+-----------+------------+

I would like to get all values of columns A,AB, Value for the latest Date on Column A filtered on AB = I

basically the result should look like:

+----+-----+----+-----------+------------+
| Id |  A  | AB |   Value   |    Date    |
+----+-----+----+-----------+------------+
|  1 | abc | I  | -48936.08 | 2013-06-24 |
|  3 | xyz | I  | -13523.90 | 2013-06-24 |
|  7 | def | I  | -13523.90 | 2013-06-22 |
+----+-----+----+-----------+------------+

I have tried to use inner join twice on the same table but failed to come up with correct result.

any help would be appreciated.

thanks :)

Answer 1

This will work with sqlserver 2005+

;WITH a as
(
SELECT id, A,AB, Value, Date
, row_number() over (partition by A order by Date desc) rn
FROM t1
WHERE AB = 'I'
)
SELECT id, A,AB, Value, Date
FROM a WHERE rn = 1

Answer 2

; WITH x AS (
  SELECT id
       , a
       , ab
       , "value"
       , "date"
       , Row_Number() OVER (PARTITION BY a ORDER BY "date" DESC) As row_num
  FROM   your_table
  WHERE  ab = 'I'
)
SELECT *
FROM   x
WHERE  row_num = 1