Напредни извештаи од базата (SQL и складирани процедури)

• Ресторан со најмногу нарачки во последните 3 месеци

  SELECT q1.restoran_ime as restoran, q1.naracki as broj_na_naracki 
  FROM(
  SELECT r.restoran_ime, COUNT(n.naracka_id) as naracki
  FROM restoran as r
  JOIN meni as m on r.restoran_id=m.restoran_id 
  JOIN obrok as o on m.meni_id=o.meni_id 
  JOIN se_sostoi_od as sso on o.obrok_id=sso.obrok_id 
  JOIN naracka as n on sso.naracka_id=n.naracka_id
  where n.naracana_na between (now()-interval '3 months') and now()
  GROUP BY r.restoran_ime) q1
  WHERE q1.naracki=(SELECT MAX(naracki)FROM (
  SELECT r.restoran_ime, COUNT (n.naracka_id) as naracki
  FROM restoran as r
  JOIN meni as m on r.restoran_id=m.restoran_id 
  JOIN obrok as o on m.meni_id=o.meni_id 
  JOIN se_sostoi_od as sso on o.obrok_id=sso.obrok_id 
  JOIN naracka as n on sso.naracka_id=n.naracka_id
  where n.naracana_na between (now()-interval '3 months') and now()
  GROUP BY r.restoran_ime) q2
  )
  

• Ресторан со најголем профит

  select rr.restoran_ime as restoran, q1.cena_vkupna as profit from restoran rr join(select r.restoran_id,sum(sso.cena * sso.kolicina) as cena_vkupna 
  from restoran r
  join meni as m on r.restoran_id=m.restoran_id 
  join obrok as o on m.meni_id=o.meni_id 
  join se_sostoi_od as sso on o.obrok_id=sso.obrok_id 
  join naracka as n on sso.naracka_id = n.naracka_id
  group by r.restoran_id)q1
  on q1.restoran_id=rr.restoran_id
  where cena_vkupna=(select max(cena_vkupna) from(
  select r.restoran_id,sum(sso.cena * sso.kolicina) as cena_vkupna 
  from restoran r
  join meni as m on r.restoran_id=m.restoran_id 
  join obrok as o on m.meni_id=o.meni_id 
  join se_sostoi_od as sso on o.obrok_id=sso.obrok_id 
  join naracka as n on sso.naracka_id = n.naracka_id
  group by r.restoran_id
  )q2)
  

• Најнарачан производ по ресторан?