Question

根据架构，Orders表具有以下列：

OrderId integer, CustomerId integer, RetailerId integer, ProductId integer, Count integer

我正在尝试找出每个零售商的订单数量最多的产品。

因此，要总结每种产品的所有订单，我有以下查询：

SELECT RetailerId, ProductId, SUM(Count) AS ProductTotal
FROM Orders
GROUP BY RetailerId, ProductId
ORDER BY RetailerId, ProductTotal DESC;

这给出了这样的输出：

RETAILERID  PRODUCTID PRODUCTTOTAL
---------- ---------- ------------
         1          5          115
         1         10           45
         1          1           15
         1          4            2
         1          8            1
         2          9           12
         2         11           10
         2          7            1
         3          3            3
         4          2            1
         5         11            1

现在，我要做的就是找到每个零售商的订单数量最多的产品。现在，它显示了所有产品；我只想要一个。

我尝试了太多的事情。以下是一个特别令人讨厌的示例：

SELECT O.RetailerId, O.ProductId, O.ProductTotal
FROM (
  SELECT Orders.ProductId, MAX(ProductTotal) AS MaxProductTotal
  FROM (
    SELECT Orders.ProductId AS PID, SUM(Orders.Count) AS ProductTotal
    FROM Orders
    GROUP BY Orders.ProductId
  ) AS O INNER JOIN Orders ON Orders.ProductId = PID
  GROUP BY Orders.ProductId
) AS X INNER JOIN O ON O.RetailerId = X.RetailerId AND O.ProductTotal = X.MaxProductTotal;

解决方案可能是有史以来最简单的事情，但是我现在不能。所以，我想要一些帮助。

Answer 1

通过窗口功能选择每个客户的最大总计：

SELECT RetailerId, ProductId, ProductTotal
FROM
(
  SELECT
    RetailerId, ProductId, SUM(Count) AS ProductTotal,
    MAX(SUM(Count)) OVER (PARTITION BY RetailerId) AS MaxProductTotal
  FROM Orders
  GROUP BY RetailerId, ProductId
)
WHERE ProductTotal = MaxProductTotal
ORDER BY RetailerId;

Answer 2

您可以尝试使用窗口功能row_number()

select * from
(
select *,row_number() over(partition by RetailerId order by ProductTotal desc) as rn from
(
SELECT RetailerId, ProductId, SUM(Count) AS ProductTotal
FROM Orders
GROUP BY RetailerId, ProductId
)A
)X where rn=1

Answer 3

您可以将with cte as ( SELECT o.RetailerId,o.ProductId AS PID, SUM(o.Count) AS ProductTotal FROM Orders o GROUP BY o.ProductId,o.RetailerId ), cte2 as ( select RetailerId,PID,ProductTotal, row_number() over(partition by RetailerId order by ProductTotal desc) as rn from cte ) select RetailerId,PID,ProductTotal from cte2 where rn=1与cte一起使用

l1 = ['A','B','C','D','A','B']
l2 = []

Answer 4

您尝试使用等级吗？

尝试此查询

select OrderC, retailer_id,product_id from ( select sum(count_order) as OrderC,retailer_id,product_id,row_number() over(order by OrderC desc) as RN from order_t  group  by retailer_id,product_id) as d where RN=1;

Answer 5

以下是使用FIRST / LAST进行子查询的版本：

WITH t(RETAILERID, PRODUCTID, PRODUCTTOTAL) AS (    
    SELECT 1,  5,  115 FROM dual UNION ALL
    SELECT 1, 10,   45 FROM dual UNION ALL
    SELECT 1,  1,   15 FROM dual UNION ALL
    SELECT 1,  4,    2 FROM dual UNION ALL
    SELECT 1,  8,    1 FROM dual UNION ALL
    SELECT 2,  9,   12 FROM dual UNION ALL
    SELECT 2, 11,   10 FROM dual UNION ALL
    SELECT 2,  7,    1 FROM dual UNION ALL
    SELECT 3,  3,    3 FROM dual UNION ALL
    SELECT 4,  2,    1 FROM dual UNION ALL
    SELECT 5,11,    1 FROM dual)
SELECT 
   RETAILERID, 
   MAX(PRODUCTID) KEEP (DENSE_RANK LAST ORDER BY PRODUCTTOTAL) AS PRODUCTID, 
   MAX(PRODUCTTOTAL)
FROM t
GROUP BY RETAILERID;


+--------------------------------------+
|RETAILERID|PRODUCTID|MAX(PRODUCTTOTAL)|
+--------------------------------------+
|1         |5        |115              |
|2         |9        |12               |
|3         |3        |3                |
|4         |2        |1                |
|5         |11       |1                |
+--------------------------------------+

查找SQL中具有最高值的行

5 个答案: