Question

我有以下名为Order

的表

Orders Table
___________________________________________________
orderiD | userId | OrderType | Order_Date | Amount
________|________|___________|____________|________
1          1          0         12/12/2009    1
2          1          1         13/12/2009    2
3          1          1         14/12/2009    3
4          2          0         12/12/2009    4
5          2          1         16/12/2009    2
6          1          0         14/12/2009    5
7          2          1         17/12/2009    4
8          2          0         10/12/2010    2
___________________________________________________

我需要创建一个返回最大购买次数的用户ID的查询。

我尝试了以下

Select MAX(GRP.sumAmmount), o.userId join
(Select SUM(o.Amount) as sum_ammount, o.userId as UID from Orders GROUP BY(o.userID)) as GRP on o.userId=GRP.UID GROUP BY(GRP.UID)

但我相信我错过了一些东西。

你能帮忙吗？

Answer 1

如果我正确理解您的问题，您希望返回具有最大购买金额（总）的UserID。因此，上述记录将产生：

UserID  Total Amount
  2         12

最简单的解决方案是：

SELECT UserID, SUM(AMOUNT) as TotalAmount
FROM Orders
GROUP BY UserID
ORDER BY TotalAmount DESC
LIMIT 1

如果我错了，我准备编辑这个。 :-)
（ PS：请添加您想要的结果）
感谢。

更新1

来自您的查询：

  Select  MAX(SUM(o.Amount)) as sum_ammount, 
          o.userId as UID 
    FROM  Orders o
GROUP BY  o.userID
ORDER BY  sum_ammount DESC
   LIMIT  1

Answer 2

见下文，其工作情况。

SELECT userId, sum(Amount) as 'Total'
FROM Orders 
GROUP BY userId 
ORDER BY Total DESC
LIMIT 1

输出

+++++++++++++++
userId + Total
+++++++++++++++
2      + 12
+++++++++++++++

我在添加新行后尝试了

insert into Orders values (9,1,0,'2010-12-10 12:12:12',10)

输出

+++++++++++++++
userId + Total
+++++++++++++++
1      + 21
+++++++++++++++

Answer 3

我更喜欢这个解决方案

SELECT UserID, SUM(AMOUNT) as TotalAmount
FROM Orders
GROUP BY UserID
having SUM(AMOUNT) = (select sum(amount) as col1
                 from orders
                 group by userid
                 order by  col1 desc
                 limit 1
                 )

此sql将显示所有拥有最大购买量的用户

尝试将orderid 1的数量更改为2，然后此sql将显示两者。

MYSQL选择最大购买总和

3 个答案: