== Напредни извештаи од базата (SQL и складирани процедури)

- Ресторан со најмногу нарачки во последните 3 месеци
{{{#!sql
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
{{{#!sql
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?