Question

我正在研究简单的查询来学习MySQL，在我的示例数据库中，我跟踪销售电子设备的商店，我有一张桌子卖（商店，物品，价格）。

示例数据是，

'Best Buy', 'Galaxy S', 1000 
'Buy More', 'Macbook Air', 2000 
'Best Buy', 'Microsoft Mouse', 20 
'Best Buy', 'Macbook Pro Cover', 40 
'Buy More', 'Asus Zenbook', 2000

等等..

我尝试了以下sql语句，但它说：

错误代码：1111。无效使用组功能0.000秒

SELECT store
FROM sells
WHERE SUM(price) <
        (SELECT SUM(price) AS total
        FROM sells
        GROUP BY store
        ORDER BY total DESC
        LIMIT 1) 
GROUP BY store
ORDER BY SUM(price) DESC

如果你能帮助我，我将不胜感激

由于

Answer 1

这只会显示第二个最贵的商店;

SELECT STORE 
FROM TABLE_A
GROUP BY STORE
ORDER BY SUM(PRICE) DESC
LIMIT 1,1

演示here。

如果您想要显示价格，您也可以选择它;

SELECT STORE, SUM(PRICE) TOTAL_PRICE
FROM TABLE_A
GROUP BY STORE
ORDER BY TOTAL_PRICE DESC
LIMIT 1,1

演示here。

编辑：如果你有几个最昂贵的商店和几个第二昂贵的商店，那么获得所有第二个最昂贵商店的查询变得相当复杂;我相信有人可以击败这个效率;

SELECT STORE, SUM(PRICE) TOTAL_PRICE
FROM TABLE_A
GROUP BY STORE
HAVING TOTAL_PRICE =
  (SELECT SUM(PRICE) TMP 
   FROM TABLE_A 
   GROUP BY STORE
   HAVING TMP <
     (SELECT SUM(PRICE) TMP2
      FROM TABLE_A
      GROUP BY STORE
      ORDER BY TMP2 DESC
      LIMIT 1)
   ORDER BY TMP DESC LIMIT 1)

演示here。

Answer 2

你可以这样做;

SELECT *,
SUM(price) AS totalprice
FROM sells
GROUP BY store 
ORDER BY totalprice DESC
LIMIT 2

首先选择价格总和并将其临时存储在ex。总价格，然后你已经逐个商店。要获得最昂贵的商店，请向后订购金额，然后将结果限制为两个。

当您循环结果时，您将能够将总价格作为普通列

Answer 3

几乎正确，

  SELECT SUM(price) as price_total FROM sells GROUP BY store

如果你想订购你可以做子查询，比如：

   SELECT price_total FROM (SELECT SUM(price) as price_total FROM sells GROUP BY store) as res ORDER BY price LIMIT 2

如果您想获得第二名，您可能会进行另一次查询，但我认为最好使用您的后端语言

Answer 4

SELECT distinct price from sells ORDER BY price DESC，在您的代码中，只需选择第二个。

如果您需要其余信息，请执行以下操作：

SELECT * from sells
WHERE price = (SELECT distinct price from sells ORDER BY price DESC LIMIT 1,1)

Answer 5

没有测试它但应该工作

SELECT S.store
FROM (
SELECT SUM(T.price) AS sum_price
FROM formList_Total AS T
GROUP BY  T.store
) AS S
ORDER BY sum_price DESC 
LIMIT 1 , 1

对不起，去了测试，这是我最终的结果。

在MySQL中找到第二大最昂贵的产品

5 个答案: