SQL Work out average from joined column - sql

Question

I have 3 columns I need to display and I need to join on another column that calculates the AVG from the CLUB_FEE column. My code does not work, it throws a "not a single-group group function" Can someone please help? Here is my SQL:
SELECT S.MEMBER_ID, S.CLUB_ID, C.CLUB_FEE, AVG(C.CLUB_FEE) AVGINCOME
FROM SUBSCRIPTION S, CLUB C
WHERE S.CLUB_ID = C.CLUB_ID;

Answer 1

i Suggest to use Inner join try it also When you include an aggregate function (like avg, sum) in your query, you must group by all columns :
SELECT S.MEMBER_ID, S.CLUB_ID, C.CLUB_FEE, AVG(C.CLUB_FEE) as AVGINCOME
FROM SUBSCRIPTION S INNER JOIN CLUB C
ON S.CLUB_ID = C.CLUB_ID
GROUP BY
S.MEMBER_ID, S.CLUB_ID, C.CLUB_FEE ;

Answer 2

Learn to use explicit JOIN syntax. Simple rule: Never use commas in the FROM clause. Always use explicit JOIN syntax.
In your case, you need to remove columns from the SELECT and the GROUP BY. If you want the average fee paid by any member, then you don't need the GROUP BY at all:
SELECT AVG(C.CLUB_FEE) as AVGINCOME
FROM SUBSCRIPTION S JOIN
CLUB C
ON S.CLUB_ID = C.CLUB_ID;
If you want to control the formatting, either use to_char():
SELECT TO_CHAR(AVG(C.CLUB_FEE), '999.99') as AVGINCOME
(check the documentation for other formats).
Or, cast to a decimal:
SELECT CAST(AVG(C.CLUB_FEE) AS DECIMAL(10, 2)) as AVGINCOME

Answer 3

If you need to display the three columns and the average, not just the average alone, you can do something like this:
SELECT S.MEMBER_ID, S.CLUB_ID, C.CLUB_FEE, A.AVGINCOME
FROM SUBSCRIPTION S INNER JOIN CLUB C
ON S.CLUB_ID = C.CLUB_ID
CROSS JOIN (SELECT AVG(CLUB_FEE) AS AVGINCOME FROM CLUB) A
;
If you need the average rounded to two decimal places, use ROUND(AVG(CLUB_FEE), 2) in the subquery.
A fancier solution, which doesn't require a join (so it doesn't scan the CLUB table twice), uses AVG as an analytic function - but doesn't partition by anything. You still need the PARTITION BY clause (with an empty column list) to indicate it's used as an analytic function, not as an aggregate.
SELECT S.MEMBER_ID, S.CLUB_ID, C.CLUB_FEE,
ROUND(AVG(C.CLUB_FEE) OVER (PARTITION BY NULL)) AS AVGINCOME
FROM SUBSCRIPTION S INNER JOIN CLUB C
ON S.CLUB_ID = C.CLUB_ID
;
Even fancier (although functionally identical) - the keyword OVER is needed to indicate analytic function, but you can also write it as OVER() (no need to even mention PARTITION BY NULL).