Question

鉴于2个表，我想从[Purchase]表中生成前3个最高金额。其他标准是[Crocs]必须包含在记录的前3位。

我有以下SQL，但它无法生成我想要的结果（结果A），请指导我如何在结果B中取出结果。谢谢。

表（购买）：

Purchase_ID | StoreID | Amount
------------|---------|--------
  1         |  21     |   22
  2         |  23     |   13
  3         |  25     |   6
  4         |  26     |   23
  5         |  28     |   18

表（商店）：

Store_ID | StoreName     
---------|----------
  21     |  Adidas
  22     |  Nike
  23     |  Puma
  24     |  New Balance
  25     |  Crocs
  26     |  Converse

SQL：

SELECT IF(SUM(amount) IS NULL, 0, SUM(amount)) as totalAmount
FROM (
    SELECT a.amount
    FROM purchase a
    INNER JOIN store b
    ON a.store_id = b.storeid
    GROUP BY a.amount
    HAVING b.StoreName = 'Crocs'
    ORDER BY a.amount DESC
    LIMIT 3
) t

结果A：$ 6

解释A：Crocs的金额为$ 6

结果B：51美元

解释B：前3名的总金额= 22美元（阿迪达斯）+ 23（美洲狮）+ 6美元（Crocs）

Answer 1

你可以使用联盟

  SELECT b.storename, max(a.amount)
  FROM purchase a
  INNER JOIN store b
  ON a.store_id = b.storeid
  GROUP BY a.storename
  order by max(a.amount) limit 2
  union 
  SELECT b.storename, a.amount
  FROM purchase a
  INNER JOIN store b
  ON a.store_id = b.storeid
  where b.storename='crocks'

Answer 2

试试这个：

https://

Answer 3

scaisEdge的答案几乎是正确的，但是第一个查询也可能返回一行带有crocs并且排序错误（按max（a.amount）限制2的顺序意味着将显示最低的2个结果）。此外，您可以将查询包装在另一个选择查询中以对结果进行排序

  SELECT * FROM (
    SELECT b.storename, max(a.amount) as maxAmount
      FROM purchase a
      INNER JOIN store b ON a.store_id = b.storeid
      WHERE b.storename != 'crocks'
      GROUP BY a.storename
      ORDER BY max(a.amount) DESC 
      LIMIT 2
    UNION 
      SELECT b.storename, a.amount as maxAmount
      FROM purchase a
      INNER JOIN store b
      ON a.store_id = b.storeid
      WHERE b.storename='crocks'
      ORDER BY a.amount DESC
      LIMIT 1
  ) ORDER BY maxAmount DESC

MySQL包含一个具有order by的特定行

3 个答案: