-- sample table:
create table Employee
(
Name varchar(1) primary key,
City varchar(10),
Age int
)
go
-- with some sample data:
insert into Employee
select 'A','Calcutta',13 union all -- odd #
select 'B','Calcutta',33 union all
select 'C','Calcutta',19 union all
select 'D','Delhi',25 union all -- single #
select 'E','Mumbai',22 union all -- even #
select 'F','Mumbai',65 union all
select 'G','Mumbai',67 union all
select 'H','Mumbai',71
go
-- here's our query, showing median age per city:
select city,
AVG(age) as MedianAge
from
(
select City, Name, Age,
ROW_NUMBER() over (partition by City order by Age ASC) as AgeRank,
COUNT(*) over (partition by City) as CityCount
from
Employee
) x
where
x.AgeRank in (x.CityCount/2+1, (x.CityCount+1)/2)
group by
x.City
go
-- clean it all up
drop table Employee
And here's the result:
city MedianAge
---------- -----------
Calcutta 19
Delhi 25
Mumbai 66
(3 row(s) affected)
Simply remove "City" from the SELECT clause and the GROUP BY clause to get the median age for all.
There may be more efficient tricks out there, but this is certainly the shortest and simplest technique I am aware of.
