Question

假设我们每个部门有3名员工。我们共有3个部门。以下是示例源表

Emp deptno salary
A    10     1000
B    10     2000
C    10     3000
D    20     7000
E    20     9000
F    20     8000
G    30     17000
H    30     15000
I    30     30000

输出

B    10     2000
F    20     8000
G    30     17000

使用分析函数dense_rank，我们可以明智地获得第二高薪。

我们可以在不使用任何分析函数的情况下实现这一目标吗？

Max（）是否也是解析函数？

Answer 1

这是一种痛苦，但你可以做到。以下查询获得第二高薪：

select t.deptno, max(t.salary) as maxs
from table t
where t.salary < (select max(salary)
                  from table t2
                  where t2.deptno = t.deptno
                 )
group by t.deptno;

然后您可以使用它来获得员工：

select t.*
from table t join
     (select t.deptno, max(t.salary) as maxs
      from table t
      where t.salary < (select max(salary)
                        from table t2
                        where t2.deptno = t.deptno
                       )
      group by t.deptno
     ) tt
     on t.deptno = tt.deptno and t.salary = tt.maxs;

Answer 2

这将为您提供每个部门的第二高薪：

SELECT Emp, deptno, salary
FROM Emp a
WHERE 1 = (SELECT COUNT(DISTINCT salary) 
        FROM Emp b 
        WHERE b.salary > a.salary AND a.deptno = b.deptno)
group by b.deptno

Answer 3

相当简单和陈述，但很慢

select 
  t1.*
from 
  #tmp t1
  inner join #tmp h1 on h1.dept = t1.dept and h1.emp <> t1.emp
  left outer join #tmp h2 on h2.dept = h1.dept and h2.salary > h1.salary
  left outer join #tmp t2 on t1.dept = t2.dept and t2.salary > t1.salary and t2.emp <> h1.emp
where 
  t2.emp is null and h2.emp is null

Answer 4

创建表格并插入虚拟数据

CREATE TABLE #Employee
(
 Id Int,
 Name NVARCHAR(10), 
 Sal int, 
 deptId int
)


INSERT INTO #Employee VALUES
(1, 'Ashish',1000,1),
(2,'Gayle',3000,1),
(3, 'Salman',2000,2),
(4,'Prem',44000,2)

查询以获得结果

 ;WITH cteRowNum AS (
SELECT *,
       DENSE_RANK() OVER(PARTITION BY deptId ORDER BY Sal DESC) AS RowNum
    FROM #Employee
 )
 SELECT *
 FROM cteRowNum
 WHERE RowNum = 2;

Answer 5

你可以找到第二高的薪水：

select max(a.Salary),a.Deptno from Employee a join (select MAX(salary) salary 
from Employee group by Deptno) b on a.Salary < b.salary group by a.Deptno

并且没有MAX（）不是分析函数。

Reference

Answer 6

在MySQL上，鉴于表名是薪水，您如何才能获得第二高的薪水：

通过嵌套查询：（您可以分别将偏移量0/1/2更改为第一，第二和第三位）

select
  *
from
  salaries as t1 
where 
  t1.salary = (select 
               salary
             from 
               salaries
             where 
             salaries.deptno = t1.deptno ORDER by salary desc limit 1 offset 1);

或者可以通过创建等级：（您可以分别将rank = 1/2/3更改为第一，第二和第三名）

SET @prev_value = NULL;
SET @rank_count = 0;
select * from 
(SELECT
  s.*, 
  CASE 
      WHEN @prev_value = deptno THEN @rank_count := @rank_count + 1
      WHEN @prev_value := deptno THEN @rank_count := 1 
      ELSE @rank_count := 1 
  END as rank
FROM salaries s
ORDER BY deptno, salary desc) as t
having t.rank = 2;

Answer 7

使用Correlated Subquery的解决方案：

SELECT * FROM emp e1 WHERE 2 = (SELECT COUNT(DISTINCT sal)
                                  FROM emp e2 
                                 WHERE e1.sal <= e2.sal
                                   AND e1.deptno = e2.deptno
                               );

Answer 8

SQL查询：

select TOP 2 max(salary),Emp from EMployee where deptno='your_detpno'

Answer 9

非常简单的逻辑。

请尝试：

SELECT dept as dd, ( SELECT ee.salary FROM `employees` as ee WHERE ee.dept=dd 
ORDER BY ee.salary DESC LIMIT 1,1 ) as sndHigh 
FROM `employees` 
WHERE 1 
GROUP BY dept

Answer 10

选择min（薪水），deptno来自（从表ORDER BY工资DESC中选择不同的前2名工资，deptno）作为一个 deptno小组

Answer 11

CREATE TABLE Employee
    ([Name] varchar(1), [Dept] varchar(1), [Salary] int)
;

INSERT INTO Employee
    ([Name], [Dept], [Salary])
VALUES
    ('a', 'M', 20),
    ('b', 'M', 25),
    ('c', 'M', 30),
    ('d', 'C', 44),
    ('e', 'C', 45),
    ('f', 'C', 46),
    ('g', 'H', 20)

如何在不使用分析功能的情况下获得第二高薪部门？

11 个答案: