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:

   PAYM_ID    USER_ID MEMBSHIP_ID PAYM_DATE                     
---------- ---------- ----------- -------------------------------
         1          1           1 18-DEC-09 12.00.00.000000000 AM 
         2          1           2 18-DEC-10 12.00.00.000000000 AM 
         3          1           2 18-DEC-11 12.00.00.000000000 AM 
         4          2           3 17-MAR-11 12.00.00.000000000 AM 
         5          3           3 18-JUN-12 12.00.00.000000000 AM 
         6          4           2 17-FEB-12 12.00.00.000000000 AM 
         7          5           2 18-FEB-11 12.00.00.000000000 AM 
         8          5           2 18-FEB-12 12.00.00.000000000 AM 
         9          6           1 01-JUN-12 12.00.00.000000000 AM 
        10          7           1 03-FEB-11 12.00.00.000000000 AM 
        11          7           2 03-FEB-12 12.00.00.000000000 AM 

I am trying with no success the following code:

SELECT MEMBSHIP_ID
FROM user_payment
WHERE user_id=1 and MAX(paym_date);

And I get this error: SQL Error: ORA-00934: group function is not allowed here 00934. 00000 - "group function is not allowed here"

How can I fix it? thanks in advance!

Answer 1

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

Answer 2

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 3

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