Question

示例表订单

+---------+----------+
| OrderID | Quantity |
+---------+----------+
|   10248 |       11 |
|   10248 |       42 |
|   10248 |       72 |
|   10249 |       14 |
|   10249 |       51 |
|   10250 |       41 |
|   10250 |       51 |
+---------+----------+

我的问题是。如何查找每个orderid的数量总和并按订单查找最高数量

<select orderid , SUM(quantity) As TotalQty  from order group by orderid

结果

+--------+-----------+
|orderId |  TotalQTY |
+--------+-----------+
|  10248 |       125 |
|  10249 |        65 |
|  10250 |        92 |
+--------+-----------+

从上面的结果我需要显示总数最高的orderid。

最终结果必须是

+--------+----------+
|orderid | TotalQTY |
+--------+----------+
|  10245 |      125 |
+--------+----------+

我在一个查询中需要这个

Answer 1

在MySQL中使用

select orderid, SUM(quantity) As TotalQty
from order 
group by orderid
order by TotalQty desc
limit 1

在MSSQL中

select top 1 orderid, SUM(quantity) As TotalQty
from order 
group by orderid
order by SUM(quantity) desc

在Oracle中

select orderid, SUM(quantity) As TotalQty
from order 
group by orderid
order by SUM(quantity) desc
where ROWNUM <= 1

Answer 2

如果是SQL-Server，您可以使用OVER clause：

SELECT TOP 1 orderid, SUM(Quantity)OVER(PARTITION BY orderid)AS TotalQTY
FROM dbo.Orders
ORDER BY TotalQTY DESC

Demo

您可以包含所有列，即使它们未汇总或属于GROUP BY。

Answer 3

Oracle：

select * from (
select orderid, SUM(quantity) As TotalQty
from order
group by orderid
order by TotalQty desc) where rownum =1;

Answer 4

MySQL的：

select orderid, SUM(quantity) As TotalQty
from order 
group by orderid
order by TotalQty desc
limit 1

Answer 5

with cte IN SQL SERVER

;with cte as
(
 select top 1 orderid,sum(TotalQTY) as x from table group by id
 )

 select * from cte
 OPTION (MAXRECURSION 1);

如何从数量总和中找到最大值

5 个答案: