My task is:
To make a query which will get Employees, who earn the biggest salary for their working experience. In other words, the Employee who earns the biggest salary with the biggest experience.
As I consider, I need to make a query with two conditions:
select * from employee where salary in (select max(salary) from employee) and
hire_date in (select min(hire_date) from employee)
I think this was what you're trying to do:
select * from (
select *,
datediff(day,hire_date,getdate()) [Days_Worked],
dense_Rank() over(Partition by datediff(day,hire_date,getdate()) order by salary desc) [RN]
from employee
)a
where a.RN = 1
order by Days_Worked DESC
So that'll give you a list of employees with the highest salary against the employees that have worked the same number of days.
Just note that with dense rank if for example there are 2 employees that have worked 88 days and both earn $50000 (higher than anyone else) it will list both employees, use ROW_NUMBER() instead of DENSE_RANK() if you want to restrict examples like that to 1 employee.
If I get it correctly, this query solves your problem.
SELECT TOP 1 WITH TIES * FROM
employee
ORDER BY
hire_date ASC,
salary DESC
Does Presto SQL really lack TOP X functionality in SELECT statements?
If so, is there a workaround in the meantime?
https://prestodb.io/
If you simply want to limit the number of rows in the result set, you can use LIMIT, with or without ORDER BY:
SELECT department, salary
FROM employees
ORDER BY salary DESC
LIMIT 10
If you want the top values per group, you can use the standard SQL row_number() window function. For example, to get the top 3 employees per department by salary:
SELECT department, salary
FROM (
SELECT department, salary row_number() OVER (
PARTITION BY department
ORDER BY salary DESC) AS rn
FROM employees
)
WHERE rn <= 3
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
I need to work out this with out using TOP
I use the following query to display top 10 salaries
SELECT Salary
from
(
SELECT Salary, Row_Number() OVER(ORDER BY SALARY DESC) AS 'Salaries'
FROM User2
)#emp
WHERE Salaries <=10
But i am getting the list as 9000,8000,7000,6000,5000,500,4000,3000,2000,10000..
1000 is missing here
What to do i tried by making
WHERE Salaries <10 (but 10000 is not displaying)
What's wrong i made can any give me the appropriate one
First off I see 500 in your list. If the salaries are not numeric, then a string sort is used. So it would be 10000, 9000,8000,7000,6000,5000,4000,3000,2000,1000. You really should change the column datatype, but if you can't you will need to use the convert function e.g. convert(numeric(9,2),Salary).
Also,I think you would be better off using the Rank function - http://msdn.microsoft.com/en-us/library/ms176102.aspx since it does what you are trying to do. Then if you need the top 15, you only need to make that change.
Example:
SELECT Salary
FROM
(SELECT Salary
,RANK() OVER
(ORDER BY SALARY DESC) AS 'RANK' From User2) Salaries
WHERE
Salaries.RANK <= 15
What's wrong with:
SELECT TOP 10 Salary
FROM User2
ORDER BY Salary DESC
You appear to be over complicating things.
Try the folowing instead.
SELECT TOP 10 Salary FROM User2 ORDER BY Salary DESC
How to find fifth highest salary in a single query in SQL Server
In SQL Server 2005 & 2008, create a ranked subselect query, then add a where clause where the rank = 5.
select
*
from
(
Select
SalesOrderID, CustomerID, Row_Number() Over (Order By SalesOrderID) as RunningCount
From
Sales.SalesOrderHeader
Where
SalesOrderID > 10000
Order By
SalesOrderID
) ranked
where
RunningCount = 5
These work in SQL Server 2000
DECLARE #result int
SELECT TOP 5 #result = Salary FROM Employees ORDER BY Salary DESC
Syntax should be close. I can't test it at the moment.
Or you could go with a subquery:
SELECT MIN(Salary) FROM (
SELECT TOP 5 Salary FROM Employees ORDER BY Salary DESC
) AS TopFive
Again, not positive if the syntax is exactly right, but the approach works.
SELECT TOP 1 salary
FROM (
SELECT DISTINCT TOP n salary
FROM employee
ORDER BY salary DESC) a
ORDER BY salary
where n > 1 -- (n is always greater than one)
You can find any number of highest salary using this query.
To find the 5th higest salary from a database, the query is..
select MIN(esal) from (
select top 5 esal from tbemp order by esal desc) as sal
its working check it out
SELECT MIN(Salary) FROM (
SELECT TOP 2 Salary FROM empa ORDER BY Salary DESC
) AS TopFive
It's working correctly, please use it.
Can be Most easily solved by:
SELECT MIN(Salary) FROM (
SELECT Salary FROM empa ORDER BY Salary DESC limit 5
)TopFive
You can try some thing like :
select salary
from Employees a
where 5=(select count(distinct salary)
from Employees b
where a.salary > b.salary)
order by salary desc
The below query to gets the Highest salary after particular Employee name.
Just have a look into that!
SELECT TOP 1 salary FROM (
SELECT DISTINCT min(salary) salary
FROM emp where salary > (select salary from emp where empname = 'John Hell')
) a
ORDER BY salary
select * from employee2 e
where 2=(select count(distinct salary) from employee2
where e.salary<=salary)
its working
You can find it by using this query:
select top 1 salary
from (select top 5 salary
from tbl_Employee
order by salary desc) as tbl
order by salary asc