select c4: vyber zenu ktora rodila ako najmaldsia:

...


and days.d.datu_od-days(d.datum_od) = 
select minimum days(d2.datum_od-
days(z2.datum_od) 
from clovek z2,clovek d2 
where z2.id_dieta=d2.id_matka)
**********************************************************
select c 5:   high level

vyberte vsetkych vdovcov

with p (id,datum ) as 
(select id_manzel,max(v.datum_od)   // hladame ku kazdemu muzovi datum                                    // posledneho manzelstva
from manzestvo group by id_manzel)
select m.id, m.meno|' '|m.priezvisko
from clovek m, manzelstvo v,clovek z,p 
where ( m.id=p.id and v.id_manzel=m.id 
		  and v.id_manzelka=z.id 
		  and v.datum_od=p.datum
		  and z.datum_do is not null 
		  and v.datum_do= z.datum_do)

**********************************************************
select c 6 : masters

vyberte vsetky dvojice osob takych ze druha je muzsky predok 
tej prvej

toto je uz taky rekurzivny prikladik
rekurzia sa robi: urob pomocnu tabulku ktora nam posluzi na rekurziu
select bude mat dve casti spojene cez union all

pohlavie nas bude zaujimat az na konci, pri vypise

with p (id_predok,id_potomok) as
select  r.id,d.id 
 from clovek r,clovek d
where (d.id_otec=r.id or d.id_matka=r.id)
union all
select predok.id,potomok.id
from p predok,clovek potomok
select c1.meno
from 

where p.id_predok=c1.id and
p.id_potomok=c2.i
and c1.pohlavie ='muz'
where (



******************************************************
level 7: ultimate beginner-pro

with  p (id,gen) as
(select id,1
from clovek
where id_otec is null
)
union all
select potomok.id,(predok.gen+1)
from p predok,clovek potomok
where predok.id= potomok.id)

select c.id,c.meno|' '|,c.priezvisko,gen
from clovek c,p
where p.id =c.id 

******************************************************
skusajte si doma:
bigamia, na cloveka napojim dve manzelstva a zistim ci maju
neprazdny prienik casu ,ak hej => ide o bigamiu

nevlastne sestry
nevlsatne sesternice
dieta pocate pred sobasom
nemanzelske deti
