在mysql中獲得每個人的第二高薪

Question

我們有如下表

person_id  |    salary
   1       |     1500
   1       |     1000
   1       |      500
   2       |     2000
   2       |     1000
   3       |     3000
   3       |     2000
   4       |     3000
   4       |     1000

我們想要每個人第二高的薪水。 按每個人分組，並獲得第二高的薪水。 像下面

person_id  |    salary
   1       |     1000
   2       |     1000
   3       |     2000
   4       |     1000

提前致謝 ：）

Answer 1

通過使用聚合函數和自連接，您可以執行以下操作

select a.*
from demo a
left join demo b on a.person_id = b.person_id
group by a.person_id,a.salary
having sum(a.salary < b.salary) = 1 /* 0 for highest 1 for second highest 2 for third and so on ... */

或使用完整的情況下表達sum

having sum(case when a.salary < b.salary then 1 else 0 end)  = 1

演示版

注意：這不能處理像一個人可能具有兩個相同的薪水值的聯系，我假設一個人的每個薪水值都將不同於一個人的其他薪水值，以處理@juergen d提到的這種情況方法將適用於其他情況聲明

Answer 2

這是一種使用exists和having子句的方法

SELECT person_id,
       Max(salary)
FROM   Yourtable a
WHERE  EXISTS (SELECT 1
               FROM   Yourtable b
               WHERE  a.person_id = b.person_id
               HAVING ( a.salary < Max(b.salary)
                        AND Count(*) > 1 )
                       OR Count(Distinct salary) = 1)
GROUP  BY person_id 

Answer 3

嘗試

select t1.*
from your_table t1
join
(
    select person_id,
           @rank := case when person_id = @prevPersonId then @rank + 1 else 1 end as rank,
           @prevPersonId := person_id
    from your_table
    cross join (select @rank := 0, @prevPersonId := 0) r
    group by person_id
    order by person_id asc, salary desc
) t2 on t1.person_id = t2.person_id
where rank = 2

Answer 4

圍繞JOIN另一種方法。

詢問

 select t1.`person_id`, max(coalesce(t2.`salary`, t1.`salary_1`)) as `salary_2` from(
   select `person_id`, max(`salary`) as `salary_1`
   from `your_table_name`
   group by `person_id`
) t1
left join `your_table_name` t2
on t1.`person_id` = t2.`person_id`
and t1.`salary_1` <> t2.`salary`
group by t1.`person_id`;

在此處查找sql​​小提琴演示

Answer 5

您也可以使用

select max(salary), person_id
from (select salary, person_id from demo 
      except 
      (select max(salary), person_id from demo group by person_id)) s
group by person_id

從初始數據集中刪除每個人的最高薪水，然后重新開始。