Select rows where every child row meets a condition - sql

Question

In my Oracle DB, I have two tables in a one-to-many relationship: Managers and Employees.
+------------+-------+------------+
| Manager_ID | Name | Department |
+------------+-------+------------+
| 1 | Steve | Sales |
| 2 | Ben | Sales |
| 3 | Molly | Accounts |
+------------+-------+------------+
+-------------+------------+--------+-----+
| Employee_ID | Manager_ID | Name | Age |
+-------------+------------+--------+-----+
| 1 | 1 | Kyle | 25 |
| 2 | 1 | Gary | 31 |
| 3 | 2 | Renee | 31 |
| 4 | 2 | Oliver | 32 |
+-------------+------------+--------+-----+
How do I select only those Managers where every one of his Employees is over the age of 30?
In my example data, the only Manager who meets this condition is Ben, because both of his employees are over 30.
I thought something like this would do it, but it's wrong:
SELECT m.manager_id
FROM managers m
WHERE m.manager_id IN (SELECT e.manager_id
FROM employees e
GROUP BY e.manager_id
HAVING e.age > 30)

Answer 1

Use not exists :
select m.*
from manager m
where not exists (select 1
from Employees e
where e.Manager_ID = m.Manager_ID and e.Age < 30
) and
exists (select 1 from Employees e where e.Manager_ID = m.Manager_ID)

Answer 2

The only thing I don't like about Yogesh's answer (which I upvoted, since it's probably the way I'd write it) is that you have to go to the employees table a second time, to make sure the manager actually has at least one employee.
On the plus side, the NOT EXISTS that Yogesh used will allow Oracle to stop looking at a manager's employees once it finds one that is too young. So, maybe it's a toss-up.
I'll offer this alternative. It is shorter than the NOT EXISTS and does not have to go to the employees table a second time.
SELECT m.*
FROM manager m
CROSS APPLY (
SELECT min(age) min_age
FROM employee e
WHERE e.manager_id = m.manager_id ) ma
where ma.min_age >= 30;

Answer 3

Using sub-query for counts
SQL> WITH manager(Manager_ID, Name, Department) AS (
2 SELECT 1, 'Steve', 'Sales' FROM dual UNION ALL
3 SELECT 2, 'Ben', 'Sales' FROM dual UNION ALL
4 SELECT 3, 'Molly', 'Accounts' FROM dual),
5 employee(Employee_ID, Manager_ID, Name, Age) AS (
6 SELECT 1 , 1, 'Kyle', 25 FROM dual UNION ALL
7 SELECT 2 ,1, 'Gary', 31 FROM dual UNION ALL
8 SELECT 3, 2, 'Renee', 31 FROM dual UNION ALL
9 SELECT 4, 2 , 'Oliver', 32 FROM dual)
10 ---------------------------
11 --- End of data preparation
12 ---------------------------
13 SELECT m.name
14 FROM manager m
15 JOIN (SELECT manager_id,
16 COUNT(1) total,
17 COUNT(CASE WHEN age > 30 THEN 1 ELSE NULL END) age_30_above
18 FROM employee
19 GROUP BY manager_id) ee
20 ON m.manager_id = ee.manager_id
21 WHERE total = age_30_above;
Output
NAME
-----
Ben
Your query will be:
SELECT m.name
FROM manager m
JOIN (SELECT manager_id,
COUNT(1) total,
COUNT(CASE WHEN age > 30 THEN 1 ELSE NULL END) age_30_above
FROM employee
GROUP BY manager_id) ee
ON m.manager_id = ee.manager_id
WHERE total = age_30_above;

Answer 4

SELECT manager_id
FROM employees -- managers
minus
select manager_id
from employees
where age <= 30

Answer 5

You can use ALL function like this:
SELECT m.manager_id
FROM managers m
WHERE (30 <= ALL (SELECT e.age FROM employees e WHERE e.manager_id = m.manager_id));

Answer 6

You might want to reverse the conditions, select all managers, who dont have any employee below 30
select * from managers
where manager_id not in (select manager_id
from employees
where age < 30)