[英]MAX(DATE) - SQL ORACLE
我想从user_id等于1的用户的表user_payments中仅选择最新的membership_id。
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
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);
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”
我该如何解决? 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
)
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);
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 +(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。 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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