Problems using GROUP BY and ORDER BY in SQL Oracle - oracle

Question

So I've been instructed to write a SQL query to display the manager number and the salary of the lowest paid employee for that manager(2 columns within a database I'm working in). I must also exclude any groups where the minimum
salary is less than or equal to $6,000. Sort the output in descending order of
salary. Now I've successfully displayed the data however, I cannot seem to order it in descending order which is what brought me here. Currently my code is as follows:
Code Snippet #1
SELECT manager_id AS "Manager ID",
MIN(salary) AS "Lowest Paid Salary"
FROM employees
WHERE manager_id IS NOT NULL
AND salary > 6000
GROUP BY manager_id;
Which gives me the following output:
Code Snippet #2
Manager ID Lowest Paid Salary
---------- ------------------
100 6500
147 6200
205 8300
108 6900
148 6100
149 6200
102 9000
101 6500
145 7000
146 7000
However, I need these values ordered from Largest -> Smallest salary. I attempted to add a GROUP BY salary DESC; At the end of the statement which would give me the following error: ORA-00979: not a GROUP BY expression
Upon some research I found that the two (GROUP BY and ORDER BY) cannot be used together in the same query as I did not include both manager_id and salary within the GROUP BY clause. So I attempted that with the following code:
SELECT manager_id AS "Manager ID",
MIN(salary) AS "Lowest Paid Salary"
FROM employees
WHERE manager_id IS NOT NULL
AND salary > 6000
GROUP BY manager_id, salary
ORDER BY salary DESC;
Which gives me a list of all the salaries for each manager_id as well as duplicates. This is not what I am looking for.
I apologize for the excess of reading, I just want to make sure you know what I've tried and what I am attempting to accomplish.
So I need to know a way I can either use both the GROUP BY and ORDER BY in the same query to display the Code snippet #2 but have it ordered. Or another way to order this data in the same format.
Thanks in advance for any help/advice.

Answer 1

You can either sort by the column number or column name
SELECT manager_id AS "Manager ID",
MIN(salary) AS "Lowest Paid Salary"
FROM #employees
WHERE manager_id IS NOT NULL
AND salary > 6000
GROUP BY manager_id
order by 2 desc
SELECT manager_id AS "Manager ID",
MIN(salary) AS "Lowest Paid Salary"
FROM #employees
WHERE manager_id IS NOT NULL
AND salary > 6000
GROUP BY manager_id
order by "Lowest Paid Salary" desc

Answer 2

You need to order by the aggregate function. ...
Try this
SELECT manager_id AS "Manager ID",
MIN(salary) AS "Lowest Paid Salary"
FROM employees
WHERE manager_id IS NOT NULL
AND salary > 6000
GROUP BY manager_id
ORDER BY min(salary) DESC;

Answer 3

Try:
order by manager_id ,salary DESC and remove the group by clause
for example
SELECT * FROM Customers
ORDER BY CustomerID DESC,CustomerName;
This statement will arrange the results according to customer id in desc order.