Question

任何人都可以通过一个查询帮助我吗？我有一个看起来像这样的数据库：表费用

paydate       receiver   payment
2011-05-06     SOLO       3000
2011-05-08     Walmart    5000
2011-05-09     McDonalds  400
2011-05-08     Korona     700
2011-05-08     Walmart    1000
2011-05-09     BigZ       1300

我必须计算最高付款完成当天所有付款的总和。结果应如下所示：

paydate       payment
2011-05-08    6700

我设法找到了最高付款额和付款额：

SELECT payment, paydate FROM expenses WHERE payment=(SELECT max(payment) FROM expenses);

但它只给了我最高付款和最后付款，我需要当天付款的总和。

Answer 1

将查询更改为：

select sum(payment), paydate
from expenses 
where paydate=(
  select paydate from expenses 
  where payment = (
    select max(payment) from expenses)
);

我使用了SUM函数。

编辑：以前的输出不是所需的。这是我以前和现在使用的sqlfiddle。

Answer 2

据我所知，其他答案没有考虑到可以有多个日期具有相同的最高付款额。我在下面的回答说明了如何返回所有可能的答案：

SELECT sum(payment) total_amount, paydate
FROM expenses
WHERE paydate in (SELECT paydate
                  From expenses
                  WHERE payment = (Select max(payment) from expenses))
group by paydate
order by total_amount desc

祝你好运！

Answer 3

CREATE TABLE `t` (
`t1` INT(10) NULL DEFAULT NULL,
`t2` DATE NULL DEFAULT NULL,
`t3` INT(10) NULL DEFAULT NULL
)


INSERT INTO `t` (`t1`, `t2`, `t3`) VALUES
(1, '2012-04-19', 100),
(2, '2012-04-18', 200),
(3, '2012-04-18', 300),
(4, '2012-04-19', 150);

和查询：

select sum(t3), t2 from t  where t2=(select t2 from t where t3 = (select max(t3) from t))

结果给出了500和2012-04-18，即总和和日期

或检查一下：

select t1, sum(t3), t2 from t  where t2=(select t2 from t where t3 = (select max(t3) from t)) group by t1

mysql总和付款

3 个答案: