[oracle] Taking the record with the max date

Let's assume I extract some set of data.

i.e.

SELECT A, date
FROM table

I want just the record with the max date (for each value of A). I could write

SELECT A, col_date
  FROM TABLENAME t_ext
 WHERE col_date = (SELECT MAX (col_date)
                     FROM TABLENAME t_in
                    WHERE t_in.A = t_ext.A)

But my query is really long... is there a more compact way using ANALYTIC FUNCTION to do the same?

You could also use:

SELECT t.*
  FROM 
        TABLENAME t
    JOIN
        ( SELECT A, MAX(col_date) AS col_date
          FROM TABLENAME
          GROUP BY A
        ) m
      ON  m.A = t.A
      AND m.col_date = t.col_date

A is the key, max(date) is the value, we might simplify the query as below:

SELECT distinct A, max(date) over (partition by A)
  FROM TABLENAME

SELECT mu_file, mudate
  FROM flightdata t_ext
 WHERE mudate = (SELECT MAX (mudate)
                     FROM flightdata where mudate < sysdate)

Since Oracle 12C, you can fetch a specific number of rows with FETCH FIRST ROW ONLY. In your case this implies an ORDER BY, so the performance should be considered.

SELECT A, col_date
FROM TABLENAME t_ext
ORDER BY col_date DESC NULLS LAST
FETCH FIRST 1 ROW ONLY;

The NULLS LAST is just in case you may have null values in your field.

If date and col_date are the same columns you should simply do:

SELECT A, MAX(date) FROM t GROUP BY A

Why not use:

WITH x AS ( SELECT A, MAX(col_date) m FROM TABLENAME )
SELECT A, date FROM TABLENAME t JOIN x ON x.A = t.A AND x.m = t.col_date

Otherwise:

SELECT A, FIRST_VALUE(date) KEEP(dense_rank FIRST ORDER BY col_date DESC)
  FROM TABLENAME
 GROUP BY A

Justin Cave answer is the best, but if you want antoher option, try this:

select A,col_date
from (select A,col_date
    from tablename 
      order by col_date desc)
      where rownum<2