| Author |
 Topic  |
|
|
aprilrocks92
Starting Member
United Kingdom
3 Posts |
 Posted - 12/03/2012 : 17:20:16
|
I have the following relational database:
employee(employee_id (primary key), salary)
manages(manager_name, employee_id (primary key))
guest(guest_id (primary key), loyalty_number, guest_name)
stays(room_id (primary key), guest_id, start_date, end_date)
I also have the following query, "Find the names of all the managers whose salary are $1000 more than the average employee salary").
The query I have written is:
select manager_name
from manages, employee
where salary > (select avg(salary) + 1000
from employee
where employee.employee_id = manages.employee_id)
However, this does not seem to give me the right answer. Any suggestions? |
Edited by - aprilrocks92 on 12/03/2012 17:47:43
|
|
|
shilpash
Yak Posting Veteran
72 Posts |
Posted - 12/03/2012 : 17:32:59
|
SELECT manager_name
FROM manages
INNER JOIN employee
ON manages.employee_id = employee.employee_id
WHERE salary > (SELECT AVG(salary) + 1000
FROM employee
) |
Edited by - shilpash on 12/03/2012 17:35:54 |
 |
|
|
aprilrocks92
Starting Member
United Kingdom
3 Posts |
Posted - 12/03/2012 : 17:36:27
|
Thank you. However, it does not seem to work. It generates the following message:
"column 'salary' does not exist".
quote:
Originally posted by shilpash
SELECT manager_name
FROM manages
WHERE salary > (SELECT AVG(salary) + 1000
FROM employee
)
|
|
|
|
shilpash
Yak Posting Veteran
72 Posts |
Posted - 12/03/2012 : 17:36:59
|
try now.I have edited the code
|
|
|
|
shilpash
Yak Posting Veteran
72 Posts |
Posted - 12/03/2012 : 17:38:19
|
Use this--
SELECT manager_name
FROM manages
INNER JOIN employee
ON manages.employee_id = employee.employee_id
WHERE salary > (SELECT AVG(salary) + 1000
FROM employee
) |
|
|
|
aprilrocks92
Starting Member
United Kingdom
3 Posts |
Posted - 12/03/2012 : 17:45:37
|
Thank you very much, this works!
quote:
Originally posted by shilpash
Use this--
SELECT manager_name
FROM manages
INNER JOIN employee
ON manages.employee_id = employee.employee_id
WHERE salary > (SELECT AVG(salary) + 1000
FROM employee
)
|
|
|
|
shilpash
Yak Posting Veteran
72 Posts |
Posted - 12/03/2012 : 17:48:58
|
| You are welcome. |
Edited by - shilpash on 12/03/2012 17:49:14 |
|
|
| |
Topic |
|
SOLVED: Simple average query SQL
