如何从派生的 table 中 SELECT 具有 MAX(value) 的行，其中值都是计算的总和？

Question

我尝试了以下代码，但我只得到了包含所有名称和总和值的完整 table，而不是包含最大值的一行：

SELECT stageName, max(total_salary)
FROM (
      SELECT c.*, sum(p.dailySalary) as total_salary        
      from contender as c
      left join participant as p
      on (p.contender = c.idContender)
      group by c.idContender ) b
group by stageName;

output: 
Yellow Jesters  205
TikTok  3073
Teabags 947
Bobbleheads 11840
Reddit  1486

但我只需要：摇头娃娃 11840

PS: 请提出一个不使用 desc 和 limit

的解决方案

Answer 1

如果您只想在结果集中排在首位，您可以排序和限制：

select c.*, sum(p.dailySalary) as total_salary        
from contender as c
left join participant as p on p.contender = c.idContender
group by c.idContender
order by total_salary desc
limit 1

如果有顶平的可能，而你又想允许，可以使用window函数：

select *
from (
    select 
        c.*, 
        sum(p.dailySalary) as total_salary, 
        rank() over(order by sum(p.dailySalary) desc) rn
    from contender as c
    left join participant as p on p.contender = c.idContender
    group by c.idContender
) t
where rn = 1

Answer 2

这是一个适用于任何版本 MySQL 5.x 的解决方案，不使用 ORDER BY、LIMIT、window 函数、视图或 CTE。

SELECT a.stagename, a.total_salary
FROM (
      SELECT c.*, sum(p.dailySalary) as total_salary        
      from contender as c
      left join participant as p
      on (p.contender = c.idContender)
      group by c.idContender ) AS a
LEFT OUTER JOIN (
      SELECT c.*, sum(p.dailySalary) as total_salary        
      from contender as c
      left join participant as p
      on (p.contender = c.idContender)
      group by c.idContender ) AS b
  ON a.total_salary < b.total_salary
WHERE b.total_salary IS NULL;

测试于 MySQL 5.7.27.

输出：

+-------------+--------------+
| stagename   | total_salary |
+-------------+--------------+
| Bobbleheads |        11840 |
+-------------+--------------+

如何从派生的 table 中 SELECT 具有 MAX(value) 的行，其中值都是计算的总和？

How can I SELECT a row with MAX(value) from a derived table where the values are all calculated sums?

mysql

sql

join

group-by

greatest-n-per-group