在 SQL 中查找具有最高值的行答案

【问题标题】：Find the row with the highest value in SQL在 SQL 中查找具有最高值的行
【发布时间】：2019-04-11 14:21:47
【问题描述】：

根据架构，Orders 表有以下列：

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

我试图找出每个零售商的订单数量最多的产品。

所以，总结每个产品的所有订单，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。

我尝试了太多东西。下面是一个特别可恶的例子：

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;

解决这个问题可能是有史以来最简单的事情，但我现在不能。所以，我需要一些帮助。

【问题讨论】：

标签： sql oracle greatest-n-per-group

【解决方案1】：

使用窗口函数选择每个客户的最大总数：

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;

【讨论】：

【解决方案2】：

您可以尝试使用窗口函数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

【讨论】：

【解决方案3】：

您可以将row_number() 与 cte 一起使用

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

【讨论】：

【解决方案4】：

你试过使用排名吗？

试试这个查询

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;

【讨论】：

【解决方案5】：

这是一个没有使用FIRST/LAST 的子查询的版本：

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                |
+--------------------------------------+

【讨论】：