Question

我有一个包含3个价格字段的数据库。现在我想按以下顺序查询最小价格

如果MIN价格1大于0则价格1

ELSE如果MIN价格2大于0则价格2

ELSE如果MIN价格3大于0则价格3

但我总是得到最低的价格，无论在哪个领域

SELECT f.user, 
(SELECT
    (
        CASE 
        WHEN MIN(s.price2) > '0' THEN CONCAT(MIN(s.price2), '*Week*') 
        WHEN MIN(s.price1) > '0' THEN CONCAT(MIN(s.price1), '*Week*')
        WHEN MIN(s.price3) > '0' THEN CONCAT(MIN(s.price3), '*Week*')
        ELSE 0
        END
    ) 
 FROM prices s WHERE f.user = s.user
) as price FROM table f WHERE f.user = 84

+------+--------+--------+--------+--------+
| id   |   user | price1 | price2 | price3 |
+------+--------+--------+--------+--------+
|    1 |     84 |    785 |      0 |      0 |
|    2 |     84 |    550 |    200 |    100 |
|    3 |     84 |      0 |     90 |     45 |
|    4 |     84 |      0 |      0 |     80 |
|    5 |     84 |      0 |    300 |     80 |
|    6 |     84 |    100 |     95 |      0 |
+------+--------+--------+--------+--------+

我想要的是：优先级为MIN price2，然后是MIN price1，然后是MIN price4。

+------+-------+-------+
| id   |  user | price |
+------+-------+-------+
|    1 |    84 |    90 |
+------+-------+-------+

Answer 1

我仍然不确定我是否理解你的问题。你拿最低价格：

MIN(price1) = 100
MIN(price2) = 90
MIN(price3) = 45

SUM(price2)具有最高优先级。因为它大于零，我们返回这个。（否则我们会检查SUM(price1)，如果它们为零或更少，我们会检查SUM(price3)。）

select
  case 
    when sum(price2) > 0 then sum(price2)
    when sum(price1) > 0 then sum(price1) end
    when sum(price3) > 0 then sum(price3) end
    else 0
  end as price
from prices
where user = 84;

MySQL SELECT MIN的优先级为字段

1 个答案: