使用SUM和GROUP BY从一列中选择MAX，而不使用子查询

Question

请考虑下表。

+----+------+--------+-------------+---------------------+
| id | user | amount | description |       paid_on       |
+----+------+--------+-------------+---------------------+
|  1 |    1 |    200 | wire        | 2017-09-01 15:45:52 |
|  2 |    2 |    200 | paypal      | 2017-09-09 05:05:05 |
|  3 |    3 |    150 | cash        | 2017-09-02 12:34:56 |
|  4 |    1 |     20 | wire        | 2017-01-09 01:23:45 |
+----+------+--------+-------------+---------------------+

我正在尝试获取每个用户的总付款以及该用户的最后付款日期。 为此，我正在使用

    SELECT
        user,
        ROUND(SUM(amount),2) AS amount,
        date_format(paid_on, '%d.%m.%Y.') AS paid_on
    FROM
        (
            SELECT
                user,
                amount,
                paid_on
            FROM payments
            WHERE paid_on BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59'
            ORDER BY paid_on DESC
         ) tmppayments
    GROUP BY user

它按照我的预期工作，因为它返回了我的想法。 但是，使用子查询似乎有点矫枉过正。 是否可以使用简单的（r）查询执行此操作，而无需借助任何子查询？

Answer 1

这会吗？

SELECT
  user,
  ROUND(SUM(amount),2) AS amount,
  date_format(MAX(paid_on),'%d.%m.%Y.') AS paid_on
FROM
  payments
GROUP BY user

请注意，我遗漏了BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59'条件，因为你从来没有提到它是你想要的东西。

Answer 2

使用MAX函数选择最新的paid_on日期。

询问

select `user`,
round(sum(`amount`),2) as `amount`,
date_format(max(`paid_on`), '%d.%m.%Y.') as `paid_on`
from `payments`
where `paid_on` between '2017-09-01 00:00:00' and '2017-09-30 23:59:59'
group by `user`;

Answer 3

然后就是不要使用子查询：

SELECT
    user,
    ROUND(SUM(amount),2) AS amount,
    MAX(date_format(paid_on, '%d.%m.%Y.')) AS max_paid_on
FROM payments
WHERE paid_on BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59'
GROUP BY user

虽然您应该知道子查询不一定会减慢您的查询速度，在这种情况下它可能不会

Answer 4

你试过这种方式吗？：

SELECT  user,
        ROUND(SUM(amount),2) AS amount,
        date_format(max(paid_on), '%d.%m.%Y.') AS paid_on
FROM payments
WHERE paid_on BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59'
GROUP BY user

Answer 5

您根本不需要子查询。 而且，您还可以简化日期逻辑：

SELECT p.user, SUM(p.amount) as total_amount, MAX(p.paid_on) as max_paid_on
FROM payments p
WHERE p.paid_on >= '2017-09-01' AND p.paid_on <'2017-10-01' 
GROUP BY user
ORDER BY max_paid_on DESC;

笔记：

• 如果您愿意，可以格式化max_paid_on 。 我更喜欢ISO标准格式，YYYY-MM-DD，以及其他任何格式。
• 我不建议将BETWEEN与日期或日期时间列一起使用。 这是一篇很好的博客文章，解释了为什么（尽管它适用于SQL Server，它确实适用于所有数据库）。
• 您的查询中存在逻辑错误。 在外部select ， paid_on未被聚合，并且不在group by 。 这几乎是任何其他数据库中的错误。
• 你对子查询是正确的。 MySQL实现了子查询，这对于不需要子查询的查询来说是不必要的开销。

Answer 6

SELECT
    p.user,
    ROUND(SUM(p.amount), 2) AS amount,
    date_format(MAX(p.paid_on), '%d.%m.%Y.') AS paid_on
FROM
    payments p
WHERE
    p.paid_on BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59'
GROUP BY
    p.user
ORDER BY
    p.paid_on DESC

Answer 7

子查询在此处被不必要地添加。 没有子查询的最佳方法是SELECT user, ROUND(SUM(amount),2) AS amount, date_format(max(paid_on), '%d.%m.%Y.') AS paid_on FROM payments WHERE paid_on BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59' GROUP BY user希望它SELECT user, ROUND(SUM(amount),2) AS amount, date_format(max(paid_on), '%d.%m.%Y.') AS paid_on FROM payments WHERE paid_on BETWEEN '2017-09-01 00:00:00' AND '2017-09-30 23:59:59' GROUP BY user帮助。