Find second lowest salary from emp table

select min(sal) from emp where sal=(select sal from emp where sal<>MIN(SAL) ) ORselect min(sal) from emp where sal=(select sal from emp having sal<>MIN(SAL) ) May this will help you.I have not tested it in SQL and I am also not sure that the syntax is exactly right .may be you need to do same changes

select min(sal) from emp where sal=(select sal from emp where sal<>MIN(SAL) )            OR                                                                                                                                  select min(sal) from emp where sal=(select sal from emp having sal<>MIN(SAL) )                                                                                                                                                 May this will help you.I have not tested it in SQL and I am also not sure that the syntax is exactly right .may be you need to do same changes
 

SELECT a.Emp_Name,a.Emp_sal FROM employee a where 2=(SELECT count(distinct(b.Emp_sal)) FROM employee b WHERE a.Emp_sal<=b.Emp_sal)

dont make simple queries the big ones,try to use as simple as possible:-

select max(sal) from emp where sal<>(select max(sal) from emp )

dont use that that is for second highest salary

sel emp_no,rank(emp_no) from emp order by acct_no asc QUALIFY RANK(emp_no) =2

select min(sal) from emp where emp.empcode in(select top 2 empcode from emp order by sal desc)

Go with this query (use inMS SQL server database)
Let me know, it is right or wrong.

select min(salary) from emp where salary in ( select distinct top 2 salary from emp order by salary desc)

select salary ,empname from emptable order by(salary) limit 1,1;

SELECT esalary FROM emp ORDER BY `esalary ` ASC LIMIT 1,1

select max(salary) from dept where salary in(select distinct top 2  salary from dept order by salary)

Tested query. Change top 2 to top 3 for 3 rd lowest and so on.

select min(sal) from emp where sal>(select max(sal) from emp );

With t1 as (select firstnme, salary, rank()over(order by salary) as rank from employee)
select * from t1 where rank=2

select min(salary) from emp where salary!=(select min(salary) from emp)

Try this it's 100% true answer

SELECT * FROM ( SELECT rownum n, id, salary FROM (select id, salary from employees order by salary )) where n = 2

This works perfectly even if you can find any nth highest or lowest salary.

Try this

SELECT min(sal) FROM emp WHERE sal > (SELECT Min(sal) FROM emp)

Inner query returns lowest salary. Then get lowest salary excluding the first lowest salary.

select * from (select rownum rank,salary from (select distinct(salary) from employees order by salary asc)) where rank = &n;

Place whatever the number that you want in place of N.

select min(sal) from (select sal from employee where sal>(select min(sal) from employee))

SELECT TOP 1 Empname,Salary FROM (SELECT TOP 2 Empname,Salary FROM Employee ORDER BY Salary ASC) X ORDER BY Salary DESC

select * from scott.emp a where 2 = ( select count(distinct(b.sal)) from scott.emp b where a.sal>=b.sal)

Select e.salary  from emp e

where 1= ( select count(*) from emp s
                  
                  where s.salary<e.salary)

 there r various query for this question

1  select min(salary)
      from emp
      where salary in ( select top 2 salary
                                   from emp
                                    order by salary desc);

SELECT * FROM

            (SELECT
                  employee_id
                  ,salary
                 ,dense_rank() OVER ( ORDER BY salary) as ranking
             FROM employees) a
WHERE a.ranking=2

select salary from emp_table order by salary asc limit 1,1;

select * from emp  a where 2=(select count(distinct(b.sal))from emp b where a.sal>=b.sal)

Select * From (Select * From sample_count_ex Where salary < (Select salary From (Select * From sample_count_ex Order By 4 Desc) Where rownum = 1) Order By 4 Desc) Where rownum = 1;

First query Select * From sample_count_ex Order By 4 Desc gets salary in descending order as we r doing order by desc;

So (Select salary From (Select * From sample_count_ex Order By 4 Desc) Where rownum = 1) gets Highest salary in that table

and (Select * From sample_count_ex Where salary < (Select salary From (Select * From sample_count_ex Order By 4 Desc) Where rownum = 1) Order By 4 Desc) gets Second highest Salary

and since rownum= 1

it returns only Second highest salary every time

select min(salary) from emp_table<>(select min(salary) from emp_table
);

Sorry Previous is not good but this one seems good.

select min(sal) from emp where sal>(select min(sal) from emp)

select sal from(select sal,rank() over (order by sal asc) as rank from emp ) where rank=2;

Select Salary From Empwhere Salary In(Select Salary From Emporder By Salary Asc where Rownum = 2);

SELECT sal FROM 

Select Sal from
(Select sal, rank() over(order by sal desc)rnk
from employee)temp where rnk=2

-- Find Second lowest Salary ----

SELECT DISTINCT(a.sal) FROM emp a WHERE 2 = ( SELECT DISTINCT(COUNT(b.sal)) FROM emp b
WHERE a.sal >= b.sal )

Code
SELECT * FROM (SELECT * FROM (SELECT DISTINCT sal FROM emp ORDER BY sal DESC) WHERE ROWNUM < n+1 ORDER BY sal) WHERE rownum<2

Code
SELECT MIN(SAL) FROM EMPLOYEE WHERE SAL>(SELECT MIN(SAL) FROM EMPLOYEE)

select max(sal) from emp where sal<>(select max(sal) from emp);

Code
SELECT * FROM emp a WHERE (&n-1) =(SELECT count(UNIQUE sal) FROM emp b WHERE a.sal>b.sal) FOR nth minimum salary
SELECT * FROM emp a WHERE (&n-1) =(SELECT count(UNIQUE sal) FROM emp b WHERE a.sal<b.sal) FOR nth maximum salary 

Ypu need to know MR. SQL that you cant use = operator with rownum ....