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MAX(DATE) - SQL ORACLE

[英]MAX(DATE) - SQL ORACLE

I want to select only the latest membership_id from table user_payments of the user with the user_id equal to 1. 我想从user_id等于1的用户的表user_payments中仅选择最新的membership_id。

This is how the table user_payment looks like: 这就是表user_payment的样子:

   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: 我尝试以下代码没有成功:

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" 我收到此错误:SQL错误:ORA-00934:此处不允许组功能00934. 00000 - “此处不允许组功能”

How can I fix it? 我该如何解决? thanks in advance! 提前致谢!

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

Try: 尝试:

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: 要么:

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

Oracle 9i+ (maybe 8i too) has FIRST/LAST aggregate functions, that make computation over groups of rows according to row's rank in group. Oracle 9i +(也许也是8i)具有FIRST / LAST聚合函数,它们根据行中的行级别对行组进行计算。 Assuming all rows as one group, you'll get what you want without subqueries: 假设所有行都是一个组,您将在没有子查询的情况下获得所需的内容:

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

试试:

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

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