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I'm using Oracle and SQL Developer. I have downloaded HR schema and need to do some queries with it. Now I'm working with table Employees. As a user, I need the sum of salary of 3 employees with highest salary in each department. I have done query for defining 3 employees with highest salary in each department:
SELECT
*
FROM
(
SELECT
employee_id,
first_name
|| ' '
|| last_name,
department_id,
salary,
ROW_NUMBER()
OVER(PARTITION BY department_id
ORDER BY
salary DESC
--ROWS BETWEEN 1 FOLLOWING AND UNBOUNDED FOLLOWING
) result
FROM
employees
)
WHERE
result <= 3;
I need to use means of window clause. I have done something like this:
SELECT
department_id,
SUM(salary)
OVER (PARTITION BY department_id ORDER BY salary
ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) total_sal
FROM
(
SELECT
employee_id,
first_name
|| ' '
|| last_name,
department_id,
salary,
ROW_NUMBER()
OVER(PARTITION BY department_id
ORDER BY
salary DESC
--ROWS BETWEEN 1 FOLLOWING AND UNBOUNDED FOLLOWING
) result
FROM
employees
)
WHERE
result <= 3;
Here is the result:
It has the necessary sum for 3 people in department and other unnnecessary results for 2 and so on. I need such result:
How can I modify my query to receive appropriate result (I need to use a window clause and analytic fuctions)?
You want aggregation rather than windowing in the outer query:
SELECT
department_id,
SUM(salary) total_sal
FROM
(
SELECT
employee_id,
first_name
|| ' '
|| last_name,
department_id,
salary,
ROW_NUMBER()
OVER(PARTITION BY department_id
ORDER BY
salary DESC
--ROWS BETWEEN 1 FOLLOWING AND UNBOUNDED FOLLOWING
) result
FROM
employees
) e
WHERE
result <= 3
GROUP BY department_id
I we were to do the same task with window functions only, then, starting from the existing query, we can either add another level of nesting of some sort, or use WITH TIES. Both pursue the same effect, which is to limit the results to one row per group.
The latter would look like:
SELECT
department_id,
SUM(salary) OVER(PARTITION BY department_id) total_sal
FROM (
SELECT e.*,
ROW_NUMBER() OVER(PARTITION BY department_id ORDER BY salary DESC) result
FROM employees e
) e
WHERE result <= 3
ORDER BY result FETCH FIRST ROW WITH TIES
While the former would phrase as:
SELECT department_id, total_sal
FROM (
SELECT e.*,
SUM(salary) OVER(PARTITION BY department_id) total_sal
FROM (
SELECT e.*,
ROW_NUMBER() OVER(PARTITION BY department_id ORDER BY salary DESC) result
FROM employees e
) e
WHERE result <= 3
) e
where result = 1
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 have a table employee
id name dept
1 bucky shp
2 name shp
3 other mrk
How can i get the name of the department(s) having maximum number of employees ? ..
I need result
dept
--------
shp
SELECT cnt,deptno FROM (
SELECT rank() OVER (ORDER BY cnt desc) AS rnk,cnt,deptno from
(SELECT COUNT(*) cnt, DEPTNO FROM EMP
GROUP BY deptno))
WHERE rnk = 1;
Assuming you are using SQL Server and each record representing an employee. So you can use window function to get the result
WITH C AS (
SELECT RANK() OVER (ORDER BY dept) Rnk
,name
,dept
FROM table
)
SELECT TOP 1 dept FROM
(SELECT COUNT(Rnk) cnt, dept FROM C GROUP BY dept) t
ORDER BY cnt DESC
With common table expressions, count the number of rows per department, then find the biggest count, then use that to select the biggest department.
WITH depts(dept, size) AS (
SELECT dept, COUNT(*) FROM employee GROUP BY dept
), biggest(size) AS (
SELECT MAX(size) FROM depts
)
SELECT dept FROM depts, biggest WHERE depts.size = biggest.size
Based on one of the answer, Let me try to explain step by step
First of all we need to get the employee count department wise. So the firstly innermost query will run
select count(*) cnt, deptno from scott.emp group by deptno
This will give result as
Now out of this we have to get the one which is having max. employee i.e. department 30.
Also please note there are chances that 2 departments have same number of employees
The second level of query is
select rank() over (order by cnt desc) as rnk,cnt,deptno from
(
select count(*) cnt, deptno from scott.emp group by deptno
)
Now we have assigned ranking to each department
Now to select rank 1 out of it. we have a simplest outer query
select * from
(
select rank() over (order by cnt desc) as rnk,cnt,deptno from
(
select count(*) cnt, deptno from scott.emp group by deptno
)
)
where rnk=1
So we have the final result where we got the department which has the maximum employees. If we want the minimum one we have to include the department table as there are chances there is a department which has no employees which will not get listed in this table
You can ignore the scott in scott.emp as that is the table owner.
The above SQL can be practised at Practise SQL online
Created table named geosalary with columns name, id, and salary:
name id salary
patrik 2 1000
frank 2 2000
chinmon 3 1300
paddy 3 1700
I tried this below code to find 2nd highest salary:
SELECT salary
FROM (SELECT salary, DENSE_RANK() OVER(ORDER BY SALARY) AS DENSE_RANK FROM geosalary)
WHERE DENSE_RANK = 2;
However, getting this error message:
ERROR: subquery in FROM must have an alias
SQL state: 42601
Hint: For example, FROM (SELECT ...) [AS] foo.
Character: 24
What's wrong with my code?
I think the error message is pretty clear: your sub-select needs an alias.
SELECT t.salary
FROM (
SELECT salary,
DENSE_RANK() OVER (ORDER BY SALARY DESC) AS DENSE_RANK
FROM geosalary
) as t --- this alias is missing
WHERE t.dense_rank = 2
The error message is pretty obvious: You need to supply an alias for the subquery.
Here is a simpler / faster alternative:
SELECT DISTINCT salary
FROM geosalary
ORDER BY salary DESC NULLS LAST
OFFSET 1
LIMIT 1;
This finds the "2nd highest salary" (1 row), as opposed to other queries that find all employees with the 2nd highest salary (1-n rows).
I added NULLS LAST, as NULL values typically shouldn't rank first for this purpose. See:
PostgreSQL sort by datetime asc, null first?
SELECT department_id, salary, RANK1 FROM (
SELECT department_id,
salary,
DENSE_RANK ()
OVER (PARTITION BY department_id ORDER BY SALARY DESC)
AS rank1
FROM employees) result
WHERE rank1 = 3
This above query will get you the 3rd highest salary in the individual department. If you want regardless of the department, then just remove PARTITION BY department_id
Your SQL engine doesn't know the "salary" column of which table you are using, that's why you need to use an alias to differentiate the two columns.
Try this:
SELECT salary
FROM (SELECT G.salary ,DENSE_RANK() OVER(ORDER BY G.SALARY) AS DENSE_RANK FROM geosalary G)
WHERE DENSE_RANK=2;
WITH salaries AS (SELECT salary, DENSE_RANK() OVER(ORDER BY SALARY) AS DENSE_RANK FROM geosalary)
SELECT * FROM salaries WHERE DENSE_RANK=2;
select level, max(salary)
from geosalary
where level=2
connect by
prior salary>salary
group by level;
In case of duplicates in salary column below query will give the right result:
WITH tmp_tbl AS
(SELECT salary,
DENSE_RANK() OVER (ORDER BY SALARY) AS DENSE_RANK
FROM geosalary
)
SELECT salary
FROM tmp_tbl
WHERE dense_rank =
(SELECT MAX(dense_rank)-1 FROM tmp_tbl
)
AND rownum=1;
Here is SQL standard
SELECT name, salary
FROM geosalary
ORDER BY salary desc
OFFSET 1 ROW
FETCH FIRST 1 ROW ONLY
To calculate nth highest salary change offset value
SELECT MAX(salary) FROM geosalary WHERE salary < ( SELECT MAX(salary) FROM geosalary )
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;