MySQL-需要找到MAX的SUM查询

Question

我需要找到预订费用最高的用户，并且如果有2个或更多用户以相同的金额显示所有预订，那么我需要MAX of SUM。

我的餐桌预订缩短了：

import csv

def code_to_names(code):
    names = []
    with open('offense_codes.csv') as csv_file:
        reader = csv.reader(csv_file, delimiter=',')
        next(reader)  # skip the first row
        for row in reader:
            names.append(line)
    return names

所以我拥有这段代码

reservation_id, user_id, performance_id, amount_to_pay, date

我知道了

SELECT user_id, SUM(amount_to_pay) FROM reservation GROUP BY user_id

它需要向用户1和用户2显示9000。

Answer 1

一种解决方案是将HAVING子句与相关子查询一起使用，该子查询获取所有max之间的sums值，并将行限制为SUM(amount_to_pay)等于max_value。

SELECT
   user_id,
   SUM(amount_to_pay) AS total
FROM
   reservation AS r
GROUP BY
   user_id
HAVING
   total = (SELECT SUM(amount_to_pay) AS tot
            FROM reservation
            GROUP BY user_id
            ORDER BY tot DESC LIMIT 1)

在线示例：DB-Fiddle

---

更新后评论：

对于sum，只有active保留，您可以采用以下一种方法：

A）添加对外部查询和子查询的限制：

SELECT
   user_id,
   SUM(amount_to_pay) AS total
FROM
   reservation AS r
WHERE
   status = "active"
GROUP BY
   user_id
HAVING
   total = (SELECT SUM(amount_to_pay) AS tot
            FROM reservation
            WHERE status = "active"
            GROUP BY user_id
            ORDER BY tot DESC LIMIT 1)

B）在CASE WHEN ... END方法内使用SUM()：

SELECT
   user_id,
   SUM(CASE WHEN status = "active" THEN amount_to_pay END) AS total
FROM
   reservation AS r
GROUP BY
   user_id
HAVING
   total = (SELECT SUM(CASE WHEN status = "active" THEN amount_to_pay END) AS tot
            FROM reservation
            GROUP BY user_id
            ORDER BY tot DESC LIMIT 1)

在线示例：DB-Fiddle