MAX(DATE) - SQL ORACLE

Question

I want to select only the latest membership_id from table user_payments of the user with the user_id equal to 1.

This is how the table user_payment looks like:

Answer 1

select * from 
  (SELECT MEMBSHIP_ID
   FROM user_payment WHERE user_id=1
   order by paym_date desc) 
where rownum=1;

Answer 2

Oracle 9i+ (maybe 8i too) has FIRST/LAST aggregate functions, that make computation over groups of rows according to row's rank in group. Assuming all rows as one group, you'll get what you want without subqueries:

SELECT
  max(MEMBSHIP_ID)
  keep (
      dense_rank first
      order by paym_date desc NULLS LAST
  ) as LATEST_MEMBER_ID
FROM user_payment
WHERE user_id=1

Answer 3

SELECT p.MEMBSHIP_ID
FROM user_payments as p
WHERE USER_ID = 1 AND PAYM_DATE = (
    SELECT MAX(p2.PAYM_DATE)
    FROM user_payments as p2
    WHERE p2.USER_ID = p.USER_ID
)

Answer 4

Try:

SELECT MEMBSHIP_ID
  FROM user_payment
 WHERE user_id=1 
ORDER BY paym_date = (select MAX(paym_date) from user_payment and user_id=1);

Or:

SELECT MEMBSHIP_ID
FROM (
  SELECT MEMBSHIP_ID, row_number() over (order by paym_date desc) rn
      FROM user_payment
     WHERE user_id=1 )
WHERE rn = 1

Answer 5

Try with:

select TO_CHAR(dates,'dd/MM/yyy hh24:mi') from (  SELECT min  (TO_DATE(a.PAYM_DATE)) as dates from user_payment a )