How retrieve other attribute with max function like max salary with name.
to retrieve max salary along with their name please help anyone.
If my understanding of the question is correct, you can use something like this:
select name, salary from table where salary = (select max(salary) from table)
With SalaryOrder AS (
Select name, salary, Row_Number() Over(Order By salary Desc) RN
From table
)
Select /*the top clause is needed as described below*/ Top (1) *
From SalaryOrder Where RN = 1
this query may return more than a record when existing people with the same salary.
To solve this you can add more orders in CTE or use Top one to pick a random record.
RANK() should be an efficient way to get name where salary is MAX. Using sub-query will also get us the same result but it will take more time than RANK() in some cases.
Query:
SELECT name, salary
FROM
(
SELECT name, salary, RANK() over(ORDER BY salary DESC) AS rnk
FROM your_table
) AS a
WHERE rnk=1
Look at the db<>fiddle with time consumption. (The time may vary for different runs)
Related
SELECT DISTINCT
employees.departmentname,
employees.firstname,
employees.salary,
employees.departmentid
FROM employees
JOIN (
SELECT MAX(salary) AS Highest, departmentID
FROM employees
GROUP BY departmentID
) departments ON employees.departmentid = departments.departmentid
AND employees.salary = departments.highest;
Why doesn't the DISTINCT work here?
I'm trying to have each department to show only once because the question is asking the highest salary in each department.
Use the ROW_NUMBER() function, as in:
select departmentname, firstname, salary, departmentid
from (
select e.*,
row_number() over(partition by departmentid, order by salary desc) as rn
from employees e
) x
where rn = 1
I'm trying to have each department to show only once because the question is asking the highest salary in each department.
Use window functions:
SELECT e.*
FROM (SELECT e.*,
ROW_NUMBER() OVER (PARTITION BY departmentID ORDER BY salary DESC) as seqnum
FROM employees e
) e
WHERE seqnum = 1;
This is guaranteed to return one row per department, even when there are ties. If you want all rows when there are ties, use RANK() instead.
Why doesn't the DISTINCT work here?
DISTINCT is not a function; it is a keyword that will eliminate duplicate rows when ALL the column values are duplicates. It does NOT apply to a single column.
The DISTINCT keyword has "worked" (i.e. done what it is intended to do) because there are no rows where all the column values are a duplicate of another row's values.
However, it hasn't solved your problem because DISTINCT is not the correct solution to your problem. For that, you want to "fetch the row which has the max value for a column [within each group]" (as per this question).
Gwen, Elena and Paula all have the same salary
and they are in the same department
Is there any direct way of using row_number() function? I want to find 2 nd highest salary
SELECT DISTINCT id
,salary
,depid
,ROW_NUMBER() OVER (
PARTITION BY depid ORDER BY salary DESC
) AS rownum
FROM emp
WHERE rownum = 2;
It gives an error, However the below code works fine.
SELECT *
FROM (
SELECT DISTINCT id
,salary
,depid
,ROW_NUMBER() OVER (
PARTITION BY depid ORDER BY salary DESC
) AS rownum
FROM emp
) AS t
WHERE t.rownum = 2;
Is any way of directly using the row_number() function as in the first option which is giving the error?
You can not use the alias name of the same query as the condition for the where clause. You also can not use windowed queries as a passing condition in the where clause.
Here is a detailed explanation Why no windowed functions in where clauses?. It is so you need another query outside the inner query and needs to write sub-query.
You can get the Nth highest salary in SQL Server from the below query.
SELECT TOP 1 salary
FROM (
SELECT DISTINCT TOP N salary
FROM <YourTableNameHere>
ORDER BY salary DESC
) AS TEMP
ORDER BY salary
This query will give you the second highest salary ? No
SELECT id
,salary
,depid
from emp
ORDER BY salary DESC
OFFSET 1 ROWS
FETCH FIRST 1 ROWS ONLY;
Well actually, it will give you the salary that is on the second position when you order the salary's from highest to lowest... So if the highest is 100 and the second highest is 100 then you will get 100 as a result. To conclude this will return a row on the second place depending on the order by clause...
This next query will give you the second highest salary :
SELECT max(id)
, salary
, max(depid)
from emp
group by salary
ORDER BY salary DESC
OFFSET 1 ROWS
FETCH FIRST 1 ROWS ONLY;
But be aware, in case you have two employees from two different departments with the same salary then it will return you the one with the higher id and it will return the higher department id which can be incorrect.
And finally this will give you one employee that has a second largest salary with correct data:
SELECT id
, salary
, depid
from emp
where id = (SELECT max(id)
from emp
group by depid, salary
ORDER BY salary DESC
OFFSET 1 ROWS
FETCH FIRST 1 ROWS ONLY);
First, you want dense_rank(), not row_number() if you want the second highest value -- ties might get in the way otherwise.
You can use an arithmetic trick:
SELECT TOP (1) WITH TIES id, salary, depid
FROM emp
ORDER BY ABS(DENSE_RANK() over (PARTITION BY depid ORDER BY salary DESC) - 2)
The "-2" is an arithmetic trick to put the "second" values highest.
That said, I would stick with the subquery because the intent in clearer.
You could use a variation on the trick that uses a TOP 1 WITH TIES in combination with an ORDER BY ROW_NUMBER
SELECT TOP 1 WITH TIES
id,
salary,
depid
FROM emp
ORDER BY IIF(2 = ROW_NUMBER() OVER (PARTITION BY depid ORDER BY salary DESC), 1, 2)
But this trick does have the disadvantage that you can't sort it by something else.
Well, not unless you wrap it in a sub-query and sort the outer query.
A test on rextester here
I prefer to use dense_rank() instead of row_number() function with CTE (common table expression) for the scenario you have mentioned. CTE is modern, easy to use and have many cool features like it is memory resident, it can be used for DUI operations, it make code easy to understand etc.
To find Nth highest salary, the CTE look like
;with findnthsalary
as
(
select empid, deptid, salary,
dense_rank() over(partition by deptid order by salary desc) salrank
from
Employee
)
select distinct id, deptid, salary
from findnthsalary
where salrank = N
I used dense_rank() because if you use row_number() it will produce the wrong result in case multiple employees have the same salary in the same department.
I am trying to find the number of employees in a table that earn exactly the maximum salary of all the employees in the table called tblPerson.
Select Max(x.[No of Employees]) as Number, x.Salary as Salary
from
(
Select Count(Id) as [No of Employees], Salary
from tblPerson
Group by Salary
Having Salary = MAX(Salary)
)x
where x.[No of Employees]=3
Now I know this is a kind of long and complex way of doing it, but I was trying to do it using a derived table. But I am getting the error:
"Column 'x.Salary' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause"
My question is, why am I getting this particular error since the main query is a simple Select statement with a where clause. Isn't it??
Mainly, aggregate functions work only with other aggregate functions or grouped by columns.
Why? Because an aggregate function needs to know the set of values to do calculation with.
In you case, the max() will want to use all the data available for the calculation and display a single result (single row) and the other column will want to be displayed row by row. So there's a conflict.
Thank you all. Every answer helped me. However, I think I found a pretty simple way to do it:
Select top 1 Count(Id) as [No of Employees], salary
from tblPerson
Group by Salary
Order by [No of Employees] DESC
select count(*) from tblPerson where salary=(select max(salary) from tblPerson)
You get the error because 'max' is an aggregation, while you have nothing to aggregate the number by.
Select Max(x.[No of Employees]) as Number, x.Salary as Salary
from
(
Select Count(Id) as [No of Employees], Salary
from tblPerson
Group by Salary
Having Salary = MAX(Salary)
)x
---------
Group by Salary -- all other items in your select statement
---------
where x.[No of Employees]=3
however, you can also use a temporary table or variable to find the persons.
To solve this via a variable, you could do the following
declare #maxSalary Decimal
set #maxSalary = (Select max(salary) from tblperson) --insert the max value into a variable
Then either aggregate the persons (or do some other logic):
Select ID from tblperson where salary = #maxSalary
The reason for not using a group by is that using a variable is more efficient, as you search the table instead of aggregating over it.
Create a CTE (RESULT), and using DENSE_RANK function, get the highest salary, together with the EmployeeID's.
The first row of the RESULT table will give the highest salary.
Using the aggregate function COUNT, get the number of Employees with the highest Salary.
with RESULT (EmployeeID, Salary, DenseRank) as
(select EmployeeID, Salary,
DENSE_RANK() over (ORDER BY SALARY DESC) AS DenseRank
from Employee)
select TOP 1 Salary,
(select COUNT(EmployeeID)
from Employee
where Salary = (select TOP 1 Salary)
from RESULT
where DenseRank = 1)
)
from RESULT
where DenseRank = 1;
I'm a novice. I have the following Employee table.
ID Name Country Salary ManagerID
I retrieved the 3rd max salary using the following.
select name , salary From (
select name, salary from
employee sort by salary desc limit 3)
result sort by salary limit 1;
How to do the same to display 3rd max salary for each country? can we use OVER (PARTITION BY country)? I tried looking in the languageManual Windowing and Analytics but I'm finding it difficult to understand. Please help!
You're definitely on the right track with windowing functions. row_number() is a good function to use here.
select name, salary
from (
select name
, salary
, row_number() over (partition by country order by salary desc) idx
from employee ) x
where idx = 3
when you order by salary, make sure that it is a numerical type, or it will not be sorted correctly.
I have a table called workers which includes a few persons by their names, their salary and their working station. The table looks something like the following:
|Name|Station|Salary|
|Kyle|1 |2200 |
|Lisa|2 |2250 |
|Mark|3 |1800 |
|Hans|4 |1350 |
This might sound like a very obvious beginner question but I cannot get it work. I would like to select the name of the person with the highest salary. Thank you for reading, and have a nice one.
Select name
from table
where salary = (select max(salary) from table)
I dont know if you want to include ties or not (if two people have the same salary and it is the max salary.
What this does is find the max salary and then uses that in the query to find all people with that salary. You will need to replace the word table with whatever your table name is.
Try this
SELECT top 1 Name
FROM tableName
ORDER BY Salary DESC
You don't mention DBMS:
select name
from table
order by salary desc
fetch first 1 rows only
If your DBMS support OLAP functions:
select name from (
select name, row_number() over (order by salary desc) as rn
from table
) where rn = 1
Try to do:
SELECT TOP name
FROM yourtable
WHERE salary = (SELECT MAX(salary) FROM yourtable)
You must use subquery to get name and highest salary:
select Name, Salary from tb_name where salary=(select max(salary) from tb_name);