Select every counted row which count-number is higher than the average count-numbers

Question

SELECT id_film, sq1.counter 
FROM(
    SELECT id_film, COUNT(*) as counter 
    FROM film_vorstellung 
    GROUP BY id_film
) as sq1 WHERE sq1.id_film = id_film

The query returns:

id_film	counter
3	1
5	1
4	1
6	2
2	3
1	1

so far, so good.

Now i want to get every id_film which counter is higher than the average counter value.. i am trying to do that for hours now. The most common syntax error is that i cant use the average method where i tried to...

Answer 1

You can do this with a CTE and a scalar query expression. Almost in plain language, first calculate the counter values per id_film and then 'get every id_film whose counter is higher than the average counter value'.

with cntrs as -- your subquery
(
 SELECT id_film, COUNT(*) as cntr 
 FROM film_vorstellung 
 GROUP BY id_film
)
select id_film, cntr 
from cntrs
where cntr > (select avg(cntr) from cntrs);

Answer 2

if you don't want to use with , an alternative solution might be like this

select y.id_film,y.cnt
  from film_vorstellung x 
JOIN  
(   select id_film,COUNT(id_film) as  cnt ,
        CAST( (SELECT DISTINCT SUM(Count(g.id_film)) OVER () as SumCount FROM film_vorstellung g GROUP BY g.id_film) as DECIMAL) /
        CAST((SELECT count(DISTINCT id_film) FROM film_vorstellung) AS DECIMAL )  avgs 
    from film_vorstellung 
group by id_film
) y
on x.id_film=y.id_film
where y.cnt >y.avgs 
group by y.id_film,y.cnt

or

SELECT id_film, sq1.counter 
FROM(
    SELECT id_film, COUNT(*) as counter 
    FROM film_vorstellung 
    GROUP BY id_film
) as sq1 WHERE sq1.id_film = id_film
group by id_film,sq1.counter 
having AVG (sq1.counter)> (SELECT AVG(X.counter) FROM ( SELECT id_film, COUNT(*) as counter  FROM film_vorstellung group by id_film )X )
go