Question

The following query seems to have a problem to calculate “ratingValue” because there is SUM(h.liked) inside a SUM.

SELECT d.itemID1 as item, 
    sum(d.sum + d.count*SUM(h.liked))/sum(d.count) as ratingValue
FROM history h, dev d 
WHERE h.userID=:id_user
    AND d.itemID1<>h.itemID 
    AND d.itemID2=h.itemID 
GROUP BY d.itemID1,h.itemID 

For a better understanding this is the original and working query (from the Slop One algorithm) :

I just substitute the “rating” table with “history” because in my case r.ratingValue is the sum of all the “like” a user has given to an itemID (=> r.ratingValue = SELECT SUM(liked) FROM history GROUP BY h.itemID ) :

SELECT d.itemID1 as item, 
    sum(d.sum + d.count*r.ratingValue)/sum(d.count) as ratingValue 
FROM rating r, dev d 
WHERE r.userID=$userID 
    AND d.itemID1<>r.itemID 
    AND d.itemID2=r.itemID 
GROUP BY d.itemID1

Answer 1

As the error message says, you cannot next aggregation functions. I think you mean:

SELECT d.itemID1 as item, 
       sum(d.sum + d.count*h.sumliked)/sum(d.count) as ratingValue
FROM (select h.userId, h.itemId, sum(h.liked) as sumliked
      from history h
      group by h.userId, h.itemId
     ) h join
     dev d 
     on h.userID=:id_user AND
        d.itemID1<>h.itemID AND
       d.itemID2=h.itemID 
GROUP BY d.itemID1

That is, you need to do the aggregation separately, in this case using a subquery. I also fixed the join syntax in your query.