Ha Seige thanks for your reply.
I tried it and it didn't work unfortunately.
I asked my friend, also an IT-student, if he could find a solution but he came up with the same answer as you did.
Now after some thinking we came with this:
- Code: Select all
SELECT avg( punten.punt ) AS gem, COUNT( punten.bier ) AS mijnPunten, bieren.id AS bierid, bieren.naam AS biernaam, merken.naam AS merknaam FROM punten, merken, bieren WHERE bieren.merk = merken.id AND bieren.id = punten.bier GROUP BY punten.bier ORDER BY `gem` DESC , mijnPunten DESC
And it worked.
But thanks anyway.
