Question

如何在MAX之前选择第1，第2或第3个值？

通常我们按顺序和限制

来做

SELECT * FROM table1
ORDER BY field1 DESC
LIMIT 2,1

但是根据我当前的查询，我不知道如何制作它......

样本表

+----+------+------+-------+
| id | name | type | count |
+----+------+------+-------+
|  1 | a    |    1 |     2 |
|  2 | ab   |    1 |     3 |
|  3 | abc  |    1 |     1 |
|  4 | b    |    2 |     7 |
|  5 | ba   |    2 |     1 |
|  6 | cab  |    3 |     9 |
+----+------+------+-------+

我正在使用此查询的最大数量为每种类型命名

SELECT 
    `table1b`.`name`
FROM
    (SELECT 
        `table1a`.`type`, MAX(`table1a`.`count`) AS `Count`
    FROM
        `table1` AS `table1a`
    GROUP BY `table1a`.`type`) AS `table1a`
        INNER JOIN
    `table1` AS `table1b` ON (`table1b`.`type` = `table1a`.`type` AND `table1b`.`count` = `table1a`.`Count`)

我希望在max（count）

之前再添加一列名称

所以结果应该是

+------+------------+
| name | before_max |
+------+------------+
| ab   |          2 | 
| b    |          1 |
| cab  |       NULL | 
+------+------------+

请询问是否有不明确的事情;）

Answer 1

根据您给定的表（测试）结构，查询必须如下：

select max_name.name,before_max.count
from 
(SELECT type,max(count) as max
FROM `test` 
group by type) as type_max
join
(select type,name,count
from test 
) as max_name on (type_max.type = max_name.type and count = type_max.max )

left join
(select type,count
from test as t1
where count != (select max(count) from test as t2 where t1.type = t2.type)
group by type
order by count desc) as before_max on(type_max.type = before_max .type)

MAX之前的MySQL SELECT值

1 个答案: