How to apply DISTINCT clause and MIN function to WHERE statement if you are selecting all the column?

Question

I need your help. I want to show all the column on a table where I can use DISTINCT clause and MIN function.

--Here is a sample code that I've tried but failed to show the expected output
SELECT * FROM TABLE  
WHERE DOCUMENT_NUMBER IN ( 
                            SELECT DISTINCT DOCUMENT_NUMBER 
                              FROM TABLE

                                          ) 
and DESIRED_DUE_DATE IN (
                          SELECT MIN(DESIRED_DUE_DATE)
                            FROM TABLE
                              GROUP BY DOCUMENT_NUMBER)
order by DOCUMENT_NUMBER desc;

Expected output should be the DOCUMENT_NUMBER are all distinct or rather no duplicate and the DESIRED_DUE_DATE of all distinct DOCUMENT_NUMBER is at minimum value while showing all the column on the table.

the above code is still showing me duplicate Document_Number .

Answer 1

It is simpler:

SELECT DOCUMENT_NUMBER, MIN(DESIRED_DUE_DATE) DESIRED_DUE_DATE
FROM TABLE
GROUP BY DOCUMENT_NUMBER
ORDER BY DOCUMENT_NUMBER desc;

GROUP BY returns distinct values of DOCUMENT_NUMBER .
If you want all the columns of the table:

SELECT t.* 
FROM TABLE t INNER JOIN (
    SELECT DOCUMENT_NUMBER, MIN(DESIRED_DUE_DATE) DESIRED_DUE_DATE
    FROM TABLE
    GROUP BY DOCUMENT_NUMBER
) g on g.DOCUMENT_NUMBER = t.DOCUMENT_NUMBER AND g.DESIRED_DUE_DATE = t.DESIRED_DUE_DATE    
ORDER BY t.DOCUMENT_NUMBER desc;

Answer 2

Your first condition is unnecessary.

Your second would work with a correlation clause instead of group by :

SELECT t.*
FROM TABLE t
WHERE t.DESIRED_DUE_DATE = (SELECT MIN(t2.DESIRED_DUE_DATE)
                            FROM TABLE t2
                            WHERE t2.DOCUMENT_NUMBER = t.DOCUMENT_NUMBER
                           )
order by t.DOCUMENT_NUMBER desc;