It's a question I got this afternoon:
There a table contains ID, Name, and Salary of Employees, get names of the second-highest salary employees, in SQL Server
Here's my answer, I just wrote it in paper and not sure that it's perfectly valid, but it seems to work:
SELECT Name FROM Employees WHERE Salary =
( SELECT DISTINCT TOP (1) Salary FROM Employees WHERE Salary NOT IN
(SELECT DISTINCT TOP (1) Salary FROM Employees ORDER BY Salary DESCENDING)
ORDER BY Salary DESCENDING)
I think it's ugly, but it's the only solution come to my mind.
Can you suggest me a better query?
Thank you very much.
To get the names of the employees with the 2nd highest distinct salary amount you can use.
;WITH T AS
(
SELECT *,
DENSE_RANK() OVER (ORDER BY Salary Desc) AS Rnk
FROM Employees
)
SELECT Name
FROM T
WHERE Rnk=2;
If Salary is indexed the following may well be more efficient though especially if there are many employees.
SELECT Name
FROM Employees
WHERE Salary = (SELECT MIN(Salary)
FROM (SELECT DISTINCT TOP (2) Salary
FROM Employees
ORDER BY Salary DESC) T);
Test Script
CREATE TABLE Employees
(
Name VARCHAR(50),
Salary FLOAT
)
INSERT INTO Employees
SELECT TOP 1000000 s1.name,
abs(checksum(newid()))
FROM sysobjects s1,
sysobjects s2
CREATE NONCLUSTERED INDEX ix
ON Employees(Salary)
SELECT Name
FROM Employees
WHERE Salary = (SELECT MIN(Salary)
FROM (SELECT DISTINCT TOP (2) Salary
FROM Employees
ORDER BY Salary DESC) T);
WITH T
AS (SELECT *,
DENSE_RANK() OVER (ORDER BY Salary DESC) AS Rnk
FROM Employees)
SELECT Name
FROM T
WHERE Rnk = 2;
SELECT Name
FROM Employees
WHERE Salary = (SELECT DISTINCT TOP (1) Salary
FROM Employees
WHERE Salary NOT IN (SELECT DISTINCT TOP (1) Salary
FROM Employees
ORDER BY Salary DESC)
ORDER BY Salary DESC)
SELECT Name
FROM Employees
WHERE Salary = (SELECT TOP 1 Salary
FROM (SELECT TOP 2 Salary
FROM Employees
ORDER BY Salary DESC) sel
ORDER BY Salary ASC)
SELECT * from Employee
WHERE Salary IN (SELECT MAX(Salary)
FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary)
FFROM employee));
Try like this..
This might help you
SELECT
MIN(SALARY)
FROM
EMP
WHERE
SALARY in (SELECT
DISTINCT TOP 2 SALARY
FROM
EMP
ORDER BY
SALARY DESC
)
We can find any nth highest salary by putting n (where n > 0) in place of 2
Example for 5th highest salary we put n = 5
How about a CTE?
;WITH Salaries AS
(
SELECT Name, Salary,
DENSE_RANK() OVER(ORDER BY Salary DESC) AS 'SalaryRank'
FROM
dbo.Employees
)
SELECT Name, Salary
FROM Salaries
WHERE SalaryRank = 2
DENSE_RANK() will give you all the employees who have the second highest salary - no matter how many employees have the (identical) highest salary.
All of the following queries work for MySQL:
SELECT MAX(salary) FROM Employee WHERE Salary NOT IN (SELECT Max(Salary) FROM Employee);
SELECT MAX(Salary) From Employee WHERE Salary < (SELECT Max(Salary) FROM Employee);
SELECT Salary FROM Employee ORDER BY Salary DESC LIMIT 1 OFFSET 1;
SELECT Salary FROM (SELECT Salary FROM Employee ORDER BY Salary DESC LIMIT 2) AS Emp ORDER BY Salary LIMIT 1;
Another intuitive way is :-
Suppose we want to find Nth highest salary then
1) Sort Employee as per descending order of salary
2) Take first N records using rownum. So in this step Nth record here is Nth highest salary
3) Now sort this temporary result in ascending order. Thus Nth highest salary is now first record
4) Get first record from this temporary result.
It will be Nth highest salary.
select * from
(select * from
(select * from
(select * from emp order by sal desc)
where rownum<=:N )
order by sal )
where rownum=1;
In case there are repeating salaries then in innermost query distinct can be used.
select * from
(select * from
(select * from
(select distinct(sal) from emp order by 1 desc)
where rownum<=:N )
order by sal )
where rownum=1;
select MAX(Salary) from Employee WHERE Salary NOT IN (select MAX(Salary) from Employee );
Simple way WITHOUT using any special feature specific to Oracle, MySQL etc.
Suppose EMPLOYEE table has data as below. Salaries can be repeated.
By manual analysis we can decide ranks as follows :-
Same result can be achieved by query
select *
from (
select tout.sal, id, (select count(*) +1 from (select distinct(sal) distsal from
EMPLOYEE ) where distsal >tout.sal) as rank from EMPLOYEE tout
) result
order by rank
First we find out distinct salaries.
Then we find out count of distinct salaries greater than each row.
This is nothing but the rank of that id.
For highest salary, this count will be zero. So '+1' is done to start rank from 1.
Now we can get IDs at Nth rank by adding where clause to above query.
select *
from (
select tout.sal, id, (select count(*) +1 from (select distinct(sal) distsal from
EMPLOYEE ) where distsal >tout.sal) as rank from EMPLOYEE tout
) result
where rank = N;
The simple way is to use OFFSET. Not only second, any position we can query using offset.
SELECT SALARY,NAME FROM EMPLOYEE ORDER BY SALARY DESC LIMIT 1 OFFSET 1 --Second largest
SELECT SALARY,NAME FROM EMPLOYEE ORDER BY SALARY DESC LIMIT 1 OFFSET 9 --For 10th largest
I think you would want to use DENSE_RANK as you don't know how many employees have the same salary and you did say you wanted nameS of employees.
CREATE TABLE #Test
(
Id INT,
Name NVARCHAR(12),
Salary MONEY
)
SELECT x.Name, x.Salary
FROM
(
SELECT Name, Salary, DENSE_RANK() OVER (ORDER BY Salary DESC) as Rnk
FROM #Test
) x
WHERE x.Rnk = 2
ROW_NUMBER would give you unique numbering even if the salaries tied, and plain RANK would not give you a '2' as a rank if you had multiple people tying for highest salary. I've corrected this as DENSE_RANK does the best job for this.
Below query can be used to find the nth maximum value, just replace 2 from nth number
select * from emp e1 where 2 =(select count(distinct(salary)) from emp e2
where e2.emp >= e1.emp)
Here I used two queries for the following scenarios which are asked during an interview
First scenario:
Find all second highest salary in the table (Second highest salary with more than
one employee )
select * from emp where salary
In (select MAX(salary) from emp where salary NOT IN (Select MAX(salary) from
emp));
Second scenario:
Find only the second highest salary in the table
select min(temp.salary) from (select * from emp order by salary desc limit 2)
temp;
There are two way to do this first:
Use subquery to find the 2nd highest
SELECT MAX(salary) FROM employees
WHERE salary NOT IN (
SELECT MAX (salary) FROM employees)
But this solution is not much good as if you need to find out the 10 or 100th highest then you may be in trouble. So instead go for window function like
select * from
(
select salary,ROW_NUMBER() over(
order by Salary desc) as
rownum from employees
) as t where t.rownum=2
By using this method you can find out nth highest salary without any trouble.
select * from emp where salary = (
select salary from
(select ROW_NUMBER() over (order by salary) as 'rownum', *
from emp) t -- Order employees according to salary
where rownum = 2 -- Get the second highest salary
)
select max(age) from yd where age<(select max(age) from HK) ; /// True two table Highest
SELECT * FROM HK E1 WHERE 1 =(SELECT COUNT(DISTINCT age) FROM HK E2 WHERE E1.age < E2.age); ///Second Hightest age RT single table
select age from hk e1 where (3-1) = (select count(distinct (e2.age)) from yd e2 where e2.age>e1.age);//// same True Second Hight age RT two table
select max(age) from YD where age not in (select max(age) from YD); //second hight age in single table
Can we also use
select e2.max(sal), e2.name
from emp e2
where (e2.sal <(Select max (Salary) from empo el))
group by e2.name
Please let me know what is wrong with this approach
SELECT *
FROM TABLE1 AS A
WHERE NTH HIGHEST NO.(SELECT COUNT(ATTRIBUTE) FROM TABLE1 AS B) WHERE B.ATTRIBUTE=A.ATTRIBUTE;
this is the simple query .. if u want the second minimum then just change the max to min and change the less than(<) sign to grater than(>).
select max(column_name) from table_name where column_name<(select max(column_name) from table_name)
SELECT name
FROM employee
WHERE salary =
(SELECT MIN(salary)
FROM (SELECT TOP (2) salary
FROM employee
ORDER BY salary DESC) )
SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee)
If you want to display the name of the employee who is getting the second highest salary then use this:
SELECT employee_name
FROM employee
WHERE salary = (SELECT max(salary)
FROM employee
WHERE salary < (SELECT max(salary)
FROM employee);
Try This one
select * from
(
select name,salary,ROW_NUMBER() over( order by Salary desc) as
rownum from employee
) as t where t.rownum=2
http://askme.indianyouth.info/details/write-a-sql-query-to-find-the-10th-highest-employee-salary-from-an-employee-table-explain-your-answer-111
Try this to get the respective nth highest salary.
SELECT
*
FROM
emp e1
WHERE
2 = (
SELECT
COUNT(salary)
FROM
emp e2
WHERE
e2.salary >= e1.salary
)
Select * from employee where salary = (Select max(salary) from employee where salary not in(Select max(salary)from employee))
Explanation :
Query 1 : Select max(salary) from employee where salary not in(Select max(salary) from employee) - This query will retrieve second highest salary
Query 2 : Select * from employee where salary=(Query 1) - This query will retrieve all the records having second highest salary(Second highest salary may have multiple records)
I think this is probably the simplest out of the lot.
SELECT Name FROM Employees group BY Salary DESCENDING limit 2;
Try this: This will give dynamic results irrespective of no of rows
SELECT * FROM emp WHERE salary = (SELECT max(e1.salary)
FROM emp e1 WHERE e1.salary < (SELECT Max(e2.salary) FROM emp e2))**
Here's a simple approach:
select name
from employee
where salary=(select max(salary)
from(select salary from employee
minus
select max(salary) from employee));
I want to post here possibly easiest solution. It worked in mysql.
Please check at your end too:
SELECT name
FROM `emp`
WHERE salary = (
SELECT salary
FROM emp e
ORDER BY salary DESC
LIMIT 1
OFFSET 1
declare
cntr number :=0;
cursor c1 is
select salary from employees order by salary desc;
z c1%rowtype;
begin
open c1;
fetch c1 into z;
while (c1%found) and (cntr <= 1) loop
cntr := cntr + 1;
fetch c1 into z;
dbms_output.put_line(z.salary);
end loop;
end;
Using this SQL, Second highest salary will get with Employee Name
Select top 1 start at 2 salary from employee group by salary order by salary desc;
Related
I have the following table:
and executing the following query
select distinct lastname, firstname, max(salary) as salary
from employees
where salary not in (select max(salary) from employees)
group by lastname, firstname
I get the following result:
I want to find all the names of the employees that have the second highest salary which in the given table are all those with 6000 salary, but the only answers I could find was the above query. I am looking for some time now another way to do so but I can't find anything.
I am using SQL Server. Does anyone have any suggestions?
You can use dense_rank() window function for this:
select * from (
select employees.*, dense_rank() over(order by salary desc) rnk from employees
) t
where rnk = 2
If you have Sql Server 2005 or higher you could use DENSE_RANK.
with a as(
select lastname,firstname,salary,
dense_rank()over(order by salary desc)r
from employees
)
select lastname,firstname,salary
from a where r=2;
You can try the below Sql query
DECLARE #SecondHighestSalary Int
SELECT TOP 1 #SecondHighestSalary = salary FROM (
SELECT TOP 2 salary
FROM employees
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
SELECT * FROM employees where salary = #SecondHighestSalary
You can try this for getting n-th highest salary, where n = 1,2,3....(int)
SELECT TOP 1 salary FROM (
SELECT TOP n salary
FROM employees
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
Hope this will help you.
I want to get 2nd highest salary from Employee table. So please help me to find out. I tried it by using below query.
Select Max(salary) from Employee;
It gives the highest salary from salary column. But I want 2nd highest salary.
select max(sal) from
(select sal,dense_rank() over (order by sal desc) dr from emp) where dr=2;
If salaries like 5000,3000,3000,2000... Dense_rank() will give you the employees having 3000 salary
SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )
This gives you second highest.
Try this
DECLARE #N int
SET #N = 2 -- Change the value here to pick a different salary rank
SELECT Salary
FROM (
SELECT row_number() OVER (ORDER BY Salary DESC) as SalaryRank, Salary
FROM Salaries
) as SalaryCTE
WHERE SalaryRank = #N
Not only 2nd Salary but you can get all your desired highest salaries (like 3rd, 5th) by this query.
select salary from(select rownum rn,salary from
(select distinct(salary) from emp order by salary desc)) x where x.rn in (1,2,3);
Just Replace your desired salary position in place of (1,2,3), like (3,5), so it will give you 3rd and 5th highest salary, like:
select salary from(select rownum rn,salary from
(select distinct(salary) from emp order by salary desc)) x where x.rn in (3,5);
Below is the example
SELECT *
FROM (
SELECT column1,
row_number() over (order by salary desc) as [rownum]
From sal_table
) t
WHERE [rownum] = 2
If there is more than one row with same salary you can use dense_rank instead of row_number()
One more:
SELECT a.salary
FROM employee a
WHERE (:n - 1) = (SELECT COUNT(1)
FROM employee b
WHERE b.salary > a.salary);
Just replace n with the nth highest that you want.
I need to write a query that will return the third highest salaried employee in the company.
I was trying to accomplish this with subqueries, but could not get the answer. My attempts are below:
select Max(salary)
from employees
where Salary not in
(select Max(salary)
from employees
where Salary not in
(select Max(salary)
from employees));
My thought was that I could use 2 subqueries to elimitate the first and second highest salaries. Then I could simply select the MAX() salary that is remaining. Is this a good option, or is there a better way to achieve this?
The most simple way that should work in any database is to do following:
SELECT * FROM `employee` ORDER BY `salary` DESC LIMIT 1 OFFSET 2;
Which orders employees by salary and then tells db to return a single result (1 in LIMIT) counting from third row in result set (2 in OFFSET). It may be OFFSET 3 if your DB counts result rows from 1 and not from 0.
This example should work in MySQL and PostgreSQL.
Edit:
But there's a catch if you only want the 3rd highest DISTINCT salary. Than you should add the DISTINCT keyword.
In case of salary list: 100, 90, 90, 80, 70.
In the above query it will produce the 3rd highest salary which is 90. But if you mean the 3rd distinct which is 80 than you should use
SELECT DISTINCT `salary` FROM `employee` ORDER BY `salary` DESC LIMIT 1 OFFSET 2;
But there's a catch, this will return you only 1 column which is Salary, because in order to operate the distinction operation, DISTINCT can only operate on a specific set of columns.
This means we should add another wrapping query to extract the employees(There can be multiple) that matches that result. Thus I added LIMIT 1 at the end.
SELECT *
FROM `employee`
WHERE
`Salary` = (SELECT DISTINCT `Salary`
FROM `employee`
ORDER BY `salary` DESC
LIMIT 1 OFFSET 2
)
LIMIT 1;
Examples can be found HERE
You can get the third highest salary by using limit , by using TOP keyword and sub-query
TOP keyword
SELECT TOP 1 salary
FROM
(SELECT TOP 3 salary
FROM Table_Name
ORDER BY salary DESC) AS Comp
ORDER BY salary ASC
limit
SELECT salary
FROM Table_Name
ORDER BY salary DESC
LIMIT 2, 1
by subquery
SELECT salary
FROM
(SELECT salary
FROM Table_Name
ORDER BY salary DESC
LIMIT 3) AS Comp
ORDER BY salary
LIMIT 1;
I think anyone of these help you.
You may try (if MySQL):
SELECT salary FROM employee ORDER BY salary DESC LIMIT 2, 1;
This query returns one row after skipping two rows.
You may also want to return distinct salary. For example, if you have 20,20,10 and 5 then 5 is the third highest salary. To do so, add DISTINCT to the above query:
SELECT DISTINCT salary FROM employee ORDER BY salary DESC LIMIT 2, 1;
SELECT Max(salary)
FROM employee
WHERE salary < (SELECT Max(salary)
FROM employee
WHERE salary NOT IN(SELECT Max(salary)
FROM employee))
hope this helped you
If SQL Server this could work
SELECT TOP (1) * FROM
(SELECT TOP (3) salary FROM employees ORDER BY salary DESC) T
ORDER BY salary ASC
As for your number of subqueries question goes it depends on your language. Check this for more information
Is there a nesting limit for correlated subqueries in Oracle?
SELECT id
FROM tablename
ORDER BY id DESC
LIMIT 2 , 1
This is only for get 3rd highest value .
You may use this for all employee with 3rd highest salary:
SELECT * FROM `employee` WHERE salary = (
SELECT DISTINCT(`salary`) FROM `employee` ORDER BY `salary` DESC LIMIT 1 OFFSET 2
);
Some DBMS's don't allow you to run several nested queries. Here is a solution that only uses 1 nested query:
SELECT salary
FROM
(
SELECT salary
FROM employees
ORDER BY salary
LIMIT 3
) as TBL1
ORDER BY salary DESC
LIMIT 1;
It should give you the desired result. It first finds the 3 largest salaries, then selects the smallest of the three (or the third one if they are equal). Here is an SQLFiddle
I found a very good explanation in
http://www.programmerinterview.com/index.php/database-sql/find-nth-highest-salary-sql/
This query should give nth highest salary
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
SELECT MAX(salary) FROM employees GROUP BY salary ORDER BY salary DESC LIMIT 1 OFFSET 2;
SELECT * FROM employee ORDER BY salary DESC LIMIT 1 OFFSET 2;
You can use nested query to get that, like below one is explained for the third max salary. Every nested salary is giving you the highest one with the filtered where result and at the end it will return you exact 3rd highest salary irrespective of number of records for the same salary.
select * from users where salary < (select max(salary) from users where salary < (select max(salary) from users)) order by salary desc limit 1
Below query will give accurate answer. Follow and give me comments:
select top 1 salary from (
select DISTINCT top 3 salary from Table(table name) order by salary ) as comp
order by personid salary
you can get any order for salary with that:
select * from
(
select salary,row_Number() over (order by salary DESC ) RN
FROM employees
)s
where RN = 3
-- put RN equal to any number of orders.
--for your question put 3
You can find Nth highest salary by making use of just one single query which is very simple to understand:-
select salary from employees e1 where N-1=(select count(distinct
salary) from employees e2 where e2.salary>e1.salary);
Here Replace "N" with number(1,2,3,4,5...).This query work properly even when where salaries are duplicate. The simple idea behind this query is that the inner subquery
count how many salaries are greater then (N-1). When we get the count then the cursor will point to that row which is N and it simply returns the salary present in that row.
SELECT salary FROM employees e1
WHERE N-1 = (SELECT COUNT(DISTINCT salary) FROM employees e2
WHERE e2.salary > e1.salary)
Here, I have solved it with a correlated nested query. It is a generalized Query so if you want to print 4th, 5th, or any number of highest salary it will work perfectly even if there are any duplicate salaries.
So, what you have to do is simply change the N value here. So, in your case, it will be,
SELECT salary FROM employees e1
WHERE 3-1 = (SELECT COUNT(DISTINCT salary) FROM employees e2
WHERE e2.salary > e1.salary)
Note that the third highest salary may be the same the the first highest salary so your current approach wouldn't work.
I would do order the employees by salary and apply a LIMIT 3 at the end of the SQL query. You'll then have the top three of highest salaries and, thus, you also have the third highest salary (if there is one, a company may have two employees and then you wouldn't have a third highest salary).
For me this query work fine in Mysql
it will return third max salary from table
SELECT salary FROM users ORDER BY salary DESC LIMIT 1 OFFSET 2;
or
SELECT salary FROM users ORDER BY salary DESC LIMIT 2,1;
select min (salary) from Employee where Salary in (Select Top 3 Salary from Employee order by Salary desc)
SELECT TOP 1 BILL_AMT Bill_Amt FROM ( SELECT DISTINCT TOP 3 NH_BL_BILL.BILL_AMT FROM NH_BL_BILL ORDER BY BILL_AMT DESC) A
ORDER BY BILL_AMT ASC
SELECT DISTINCT MAX(salary) AS max
FROM STAFF
WHERE salary IN
(SELECT salary
FROM STAFF
WHERE salary<(SELECT MAX(salary) AS maxima
FROM STAFF
WHERE salary<
(SELECT MAX(salary) AS maxima
FROM STAFF))
GROUP BY salary);
I have tried other ways they are not right. This one works.
We can find the Top nth Salary with this Query.
WITH EMPCTE AS (
SELECT E.*, DENSE_RANK() OVER(ORDER BY SALARY DESC) AS DENSERANK
FROM EMPLOYEES E
)
SELECT * FROM EMPCTE WHERE DENSERANK=&NUM
for oracle it goes like this:
select salary from employee where rownnum<=3 order by salary desc
minus
select salary from employee where rownnum<=2 order by salary desc;
The SQL-Server implementation of this will be:
SELECT SALARY FROM EMPLOYEES OFFSET 2 ROWS FETCH NEXT 1 ROWS ONLY
This is a MYSQL query.
Explanation: The subquery returns top 3 salaries. From the returned result, we select the minimum salary, which is the 3rd highest salary.
SELECT MIN(Salary)
FROM (
SELECT Salary
FROM Employees
ORDER BY Salary DESC
LIMIT 3
) AS TopThreeSalary;
in Sql Query you can get nth highest salary
select * from(
select empname, sal, dense_rank()
over(order by sal desc)r from Employee)
where r=&n;
To find to the 2nd highest sal set n = 2
To find 3rd highest sal set n = 3 and so on.
This works fine with Oracle db.
select SAL from ( SELECT DISTINCT SAL FROM EMP ORDER BY SAL DESC FETCH FIRST 3 ROWS ONLY ) ORDER BY SAL ASC FETCH FIRST 1 ROWS ONLY
SELECT *
FROM maintable_B7E8K
order by Salary
desc limit 1 offset 2;
--Oracle SQL
with temp as (
select distinct salary from HR.EMPLOYEES
order by SALARY desc
)
select min(temp.salary) from temp
where rownum <= 3;
SELECT * FROM(
SELECT salary, DENSE_RANK()
OVER(ORDER BY salary DESC)r FROM Employee)
WHERE r=&n;
To find the 3rd highest salary set n = 3
Recently in an interview I was asked to write a query where I had to fetch nth highest salary from a table without using TOP and any sub-query ?
I got totally confused as the only way I knew to implement it uses both TOP and sub-query.
Kindly provide its solution.
Thanks in advance.
Try a CTE - Common Table Expression:
WITH Salaries AS
(
SELECT
SalaryAmount, ROW_NUMBER() OVER(ORDER BY SalaryAmount DESC) AS 'RowNum'
FROM
dbo.SalaryTable
)
SELECT
SalaryAmount
FROM
Salaries
WHERE
RowNum <= 5
This gets the top 5 salaries in descending order - you can play with the RowNumn value and basically retrieve any slice from the list of salaries.
There are other ranking functions available in SQL Server that can be used, too - e.g. there's NTILE which will split your results into n groups of equal size (as closely as possible), so you could e.g. create 10 groups like this:
WITH Salaries AS
(
SELECT
SalaryAmount, NTILE(10) OVER(ORDER BY SalaryAmount DESC) AS 'NTile'
FROM
dbo.SalaryTable
)
SELECT
SalaryAmount
FROM
Salaries
WHERE
NTile = 1
This will split your salaries into 10 groups of equal size - and the one with NTile=1 is the "TOP 10%" group of salaries.
;with cte as(
Select salary,
row_number() over (order by salary desc) as rn
from salaries
)
select salary
from cte
where rn=#n
(or use dense_rank in place of row_number if you want the nth highest distinct salary amount)
Select *
From Employee E1
Where
N = (Select Count(Distinct(E2.Salary)) From Employee E2 Where E2.Salary >= E1.Salary)
with cte as(
select VendorId,IncomeDay,IncomeAmount,
Row_Number() over ( order by IncomeAmount desc) as RowNumber
from DailyIncome
)
select * from cte
where RowNumber=2
Display 5th Min Sal Emp Table.
SELECT * FROM (SELECT Dense_Rank () Over (ORDER BY Sal ASC) AS Rnk, Emp.* FROM Emp) WHERE
Rnk=5;
Try this.
SELECT * FROM
(SELECT Salary,
rownum AS roworder
FROM (select distinct Salary from employer)
ORDER BY Salary
)
where roworder = 6
;
It can simply be done as following for second highest-
Select MAX(Salary) from employer where Salary NOT IN(Select MAX(Salary) from employer);
But for Nth highest we have to use CTE(Common Table Expression).
try this. It may very easy to find nth rank items by using CTE
**
with result AS
(
SELECT *,dense_rank() over( order by Salary) as ranks FROM Employee
)
select *from RESULT Where ranks = 2
**
To find the Nth highest salary :
Table name - Emp
emplyee_id salary
1 2000
2 3000
3 5000
4 8000
5 7000
6 2000
7 1000
sql query -> here N is higest salary to be found :
select salary from (select salary from Emp order by salary DESC LIMIT N) AS E order by ASC LIMIT 1;
If there are duplicate entries of
30,000,
23,000,
23,000,
15,000,
14,800
then above selected query will not return correct output.
find correct query as below:
with salaries as
(
select Salary,DENSE_RANK() over (order by salary desc) as 'Dense'
from Table_1
)
select distinct salary from salaries
where dense=3
SELECT salery,name
FROM employ
ORDER BY salery DESC limit 1, OFFSET n
with CTE_name (salary,name)
AS
( row_num() over (order by desc salary) as num from tablename )
select salary, name from CTE_name where num =1;
This will work in oracle
In order to find the Nth highest salary, we are only considering unique salaries.Highest salary means no salary is higher than it, Second highest means only one salary is higher than it, 3rd highest means two salaries are higher than it,similarly,Nth highest salary means N-1 salaries are higher than it.
Well, you can do by using LIMIT keyword, which provides pagination
capability.You can do like below:
SELECT salary FROM Employee ORDER BY salary DESC LIMIT N-1, 1
Ex: 2nd highest salary in MySQL without subquery:
SELECT salary FROM Employee ORDER BY salary DESC LIMIT 1,1
6- ways to write Second Highest salary..**
1.select * from employee order by Salary desc offset 1 rows fetch next 1 row only
2.select max(salary) from Employee where salary<(select max(salary) from Employee)
3.select MAX(Salary) from Employee WHERE Salary NOT IN (select MAX(Salary) from Employee );
4.select max(e1.salary) from Employee e1,Employee e2 where e1.salary
5.with cte as(
SELECT *, ROW_NUMBER() OVER( order by SALARY desc) AS ROWNUM FROM EMPLOYEE as rn)
select *From cte where ROWNUM=2
6.select max(e1.Salary) from Employee e1,Employee e2 where e1.Salary
Correct way to get nth Highest salary using NTILE function.
SELECT DISTINCT SAL INTO #TEMP_A FROM EMPLOYEE
DECLARE #CNT INT
SELECT #CNT=COUNT(1) FROM #TEMP_A
;WITH RES
AS(
SELECT SAL,NTILE(#CNT) OVER (ORDER BY SAL DESC) NTL
FROM #TEMP_A )
SELECT SALFROM RES WHERE NTL=3
DROP TABLE #TEMP_A
salary ---> table name
SELECT salary
FROM salary S1
WHERE 5-1 = (
SELECT COUNT( DISTINCT ( S2.salary ) )
FROM salary S2
WHERE S2.salary > S1.salary );
Highest sal using ms sql server:
select sal from emp where sal=(select max(sal) from emp)
Second highest sal:
select max(sal) from emp where sal not in (select max(sal) from emp)
What if we are required to find Nth highest salary without Row_Number,Rank, Dense Rank and Sub Query?
Hope this below Query Helps out.
select * from [dbo].[Test] order by salary desc
Emp_Id Name Salary Department
4 Neelu 10000 NULL
2 Rohit 4000 HR
3 Amit 3000 OPS
1 Rahul 2000 IT
select B.Salary from TEst B join Test A
on B.Salary<=A.Salary
group by (B.Salary)
having count(B.salary)=2
Result:- 4000, The 2nd Highest.
what is the query to return Name and Salary of employee Having Max Salary
SELECT Name, Salary FROM Minions
WHERE Salary = (SELECT Max(Salary) FROM Minions)
Note that this will return more than one row if there is more than one employee who have the same max salary
select name, salary from (select * from salary_table order by salary desc limit 1)
SELECT FirstName, max(salary)
FROM Employees
WHERE salary = (
SELECT max(salary)
FROM employees
)
GROUP BY FirstName
working in SQL SERVER 2012
A couple of proprietary solutions
SELECT TOP 1 [WITH ties] Name, Salary
FROM employee
ORDER BY Salary DESC
SELECT Name, Salary
FROM employee
ORDER BY Salary DESC
LIMIT 1
And a standard one
WITH E AS
(
SELECT Name, Salary,
ROW_NUMBER() OVER (ORDER BY Salary DESC) RN /*Or RANK() for ties*/
FROM employee
)
SELECT Name, Salary FROM E WHERE RN=1
In case there are multiple rows with the same MAX number then you can just limit the number to desired number like this
SELECT Name, Salary FROM Minions
WHERE Salary = (SELECT Max(Salary) FROM Minions) LIMIT 1
Select e.name, e.salary from employee e where
not exists (select salary from employee e1 where e.salary < e1.salary)
This will of course return more than one record if there are multiple people with the max salary.
If you are using an oracle database and only want a single "employee" then:
SELECT MAX( name ) KEEP ( DENSE_RANK LAST ORDER BY salary ASC ) AS name,
MAX( salary ) KEEP ( DENSE_RANK LAST ORDER BY salary ASC ) AS salary
FROM Minions;
SQLFIDDLE
(kudos to Neil N for his table name)
SQL Server has a similar FIRST_VALUE (or LAST_VALUE) analytic function.
PostgreSQL also supports window functions including LAST_VALUE.
These type of queries (grouped operaions) can execute with sub query.
Example
select *from emp where sal=(max(sal)from emp)
Assuming only one employee has maximum salary
SQL Server:
select top 1 name, salary
from employee
order by salary desc
Oracle:
select name, salary
from employee
order by salary desc
limit 1
Above queries scans the table only once unlike other queries using subqueries.