SQL subqueries multiple value - sql

Question

I just need a little help about SQL queries
Here's the situation
ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01 Jason Martin 25-JUL-96 25-JUL-06 1234.56 Toronto Programmer
02 Alison Mathews 21-MAR-76 21-FEB-86 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL> -- GROUP BY clause and AVG() function
SQL> SELECT city, AVG(salary)
2 FROM employee
3 GROUP BY city;
CITY AVG(SALARY)
---------- -----------
New York 6110.28
Toronto 1234.56
Vancouver 3823.78
The problem is I can't find a way to extract those names having higher salaries from each avg(salary) for city
Example:
Vancouver has avg(salary) of 3823.78 so I should get the name of 2 people: alison and james because they have higher salary than the avg(salary) of new york
For now I only go to this query but not working
select FIRST_NAME,SALARY,CITY
from employee
where SALARY > ( select avg(SALARY)
from employee
group by CITY
);
it tells me that subquery return more than 1 value
Hope someone can help me

Answer 1

Your query is returning more than one value. To restrict it to one value, you need a condition on city:
select FIRST_NAME,SALARY,CITY
from employee e
where SALARY > (select avg(SALARY)
from employee e2
where e2.city = e.city
group by CITY);
This is called a correlated subquery. You can also write it as a join, which is the syntax that I would use. The group by CITY is, strictly speaking, unnecessary, but I think it makes the intention of the subquery clearer.
You can rewrite this as a join by doing:
select e.FIRST_NAME, e.SALARY, e.CITY
from employee e join
(select city, avg(salary) as avgSalary
from employee
group by city
) ec
on e.city = ec.city
where e.salary > ec.avgSalary
Or, what would be my favorite way . . .
select e.FIRST_NAME, e.SALARY, e.CITY
from (select e.*,
avg(salary) over (partition by city) as avgSalary
employee e
) e
where e.salary > e.avgSalary