Question

我有一个名为'Sales`的大表，有这些区域

Time  | Product  | amount | customerid
Date1 | Product1 | 2      | 2
Date1 | Product2 | 5      | 3
Date1 | Product1 | 6      | 3
Date1 | Product4 | 1      | 3
Date2 | Product1 | 2      | 1
Date2 | Product1 | 2      | 4
Date2 | Product3 | 3      | 4
Date2 | Product3 | 3      | 1

现在，我希望每天都能获得最受欢迎的产品。

SELECT  Time, 
        Product, 
        SUM(amount) AS Total 
FROM Sales 
     GROUP BY Time, Product

给我以下

Time  | Product  | total
Date1 | Product1 | 8
Date1 | Product2 | 5
Date1 | Product4 | 1
Date2 | Product1 | 4
Date2 | Product4 | 6

我必须改变我的查询，才能获得每天最畅销的产品？

编辑：非常感谢您的解决方案，但它不起作用。我想总结一下，哪个产品是每天最畅销的产品


   Time   | Product  | total

    Date1 | Product1 | 8

    Date2 | Product4 | 6

像'LIMIT 1'这样的东西只显示第一行，但不是全部！

Answer 1

尝试此查询。 Rank功能的实现在不同的DB中是不同的。

SELECT TIME, PRODUCT
FROM (
SELECT TIME, PRODUCT, RANK() OVER(ORDER BY SUM_AMOUNT) AS SELECT_TOP
FROM 
(
(
      SELECT 
      TIME
     , PRODUCT
     , SUM(AMOUNT) AS SUM_AMOUNT
     FROM Sales 
GROUP BY TIME, PRODUCT ) TBL1 ) TBL2
WHERE
SELECT_TOP = 1

Answer 2

SELECT Time, Product, SUM(amount) AS Total FROM Sales GROUP BY Time, Product order by sum(amount) DESC;

Answer 3

SELECT Time, Product, SUM(amount) AS Total FROM Sales GROUP BY Time, Product ORDER BY Total DESC limit 1

我使用ORDER BY Total DESC按降序对Total列进行排序，以便Total列的最大值位于顶部。因此，我只需按limit 1

取最大的一个

Answer 4

查询：

SELECT 
       Time, Product, SUM(amount) AS Total 
FROM Sales 
       GROUP BY Time, Product 
       ORDER BY time Desc, Total DESC

SQL：如何获取表的每个最大值？

4 个答案: