与用户[重复]一起每月查找最大金额的SQL查询

Question

此问题已经在这里有了答案：

SQL select only rows with max value on a column [duplicate]（27个答案）

已关闭1小时前。

我想与已完成最大金额的用户一起每月打印最大金额。我的表结构是

+--------------+-------------+------+-----+---------+-------+
| Field        | Type        | Null | Key | Default | Extra |
+--------------+-------------+------+-----+---------+-------+
| customername | varchar(20) | YES  |     | NULL    |       |
| processed_at | date        | YES  |     | NULL    |       |
| amount       | int(11)     | YES  |     | NULL    |       |
+--------------+-------------+------+-----+---------+-------+

和要打印最大月总和的查询是

SELECT MAX(A.AMT), A.month
FROM ( SELECT customername,SUM(amount) AS AMT, EXTRACT( month from processed_at) as month
FROM payments
GROUP BY customername,month) AS A
GROUP BY  2;

这就是结果

| MAX(A.AMT) | month |
+------------+-------+
|       1900 |     4 |
|       2400 |     3 |

有没有一种方法可以将客户名也映射到最大金额？

Answer 1

如果您正在运行MySQL 8.0，则可以为此使用窗口函数：

SELECT *
FROM (
    SELECT 
        customername,
        SUM(amount) amt, 
        DATE_FORMAT(processed_at, '%Y-%m-01') yr_month,
        RANK() OVER(PARTITION BY DATE_FORMAT(processed_at, '%Y-%m-01') ORDER BY SUM(amount) DESC) rn
    FROM payments
    GROUP BY customername, yr_month
) t
WHERE rn = 1

请注意，如果您的数据分布在12个月以上，则GROUP BY子句中应包含月份和年。

在较早的版本中，一个选项是使用HAVING子句和相关的聚合查询进行过滤：

SELECT 
    customername,
    SUM(amount) amt, 
    DATE_FORMAT(processed_at, '%Y-%m-01') yr_month
FROM payments p
GROUP BY customername, yr_month
HAVING COUNT(*) = (
    SELECT SUM(p1.amount)
    FROM payments p
    WHERE DATE_FORMAT(p1.processed_at, '%Y-%m-01') = yr_month
    GROUP BY p1.customername
    ORDER BY SUM(p1.amount) DESC
    LIMIT 1
)