Zivjo
jaz imam takole:
7)
*/* ce je razlika prazna mnozica, je J# vsebovan */*
SELECT J.J#
FROM J
WHERE NOT EXISTS(
*/* vsi deli, ki jih prodaja trgovec X */*
SELECT DISTINCT SPJ.P#
FROM SPJ
WHERE SPJ.S#='&x'
MINUS
*/* vsi deli, ki jih prodaja trgovec X in jih kupuje projekt */*
SELECT DISTINCT SPJ.P#
FROM SPJ
WHERE SPJ.S#='&x' AND SPJ.J#=J.J#
)
upam da ti bo kaj pomagalo...
LP Blaž
On 2/14/06, Matjaž Horvat <[EMAIL PROTECTED]> wrote:
>
> Hvala.
>
> Ampak a je ta 7. ziher prav?
>
> LP,
> Matjaz
>
> rider pravi:
> >
> >
> > Matjaž Horvat pravi:
> >> Helou!
> >>
> >> Mi kdo lahko poslje resitev 7. naloge v 3. vajah, ki smo jih resevali
> >> na vajah za PB v 1. semestru?
> >>
> >> To je naloga:
> >> "Pre�itaj ťifro trgovca (X); poiť�i ťifre projektov, ki kupujejo
> >> vse potrebne dele tudi pri trgovcu X!"
> >>
> >> LP,
> >> Matjaz
> >>
> >>
> >> __________ NOD32 1.1407 (20060213) Information __________
> >>
> >> This message was checked by NOD32 antivirus system.
> >> http://www.eset.com
> >>
> >>
> >>
> >
> > ------------------------------------------------------------------------
> >
> > VAJA3
> >
> > 1.
> > SELECT S.CITY,SPJ.P#,J.CITY
> > FROM S , SPJ , J
> > WHERE S.CITY <> J.CITY AND SPJ.S#=S.S# AND SPJ.J#=J.J#;
> >
> > 2.
> > SELECT DISTINCT T1.S#
> > FROM SPJ T1
> > WHERE NOT EXISTS ((SELECT J#
> > FROM J)
> > MINUS
> > (SELECT T2.J#
> > FROM SPJ T2
> > WHERE T2.P#=T1.P# AND T2.S#=T1.S#));
> >
> > 3.
> > SELECT T.J#
> > FROM SPJ T
> > WHERE T.S#='&X' AND T.J# NOT IN (SELECT K.J#
> > FROM SPJ K
> > WHERE K.S#
> <>'&X');
> >
> > 4.
> > SELECT DISTINCT P#
> > FROM P
> > WHERE NOT EXISTS ((SELECT J.J#
> > FROM J
> > WHERE J.CITY='&X')
> > MINUS
> > (SELECT J#
> > FROM SPJ
> > WHERE SPJ.P#=P.P#));
> >
> > 5.
> > SELECT DISTINCT K.J#
> > FROM SPJ K
> > WHERE NOT EXISTS ((SELECT K1.P#
> > FROM SPJ K1
> > WHERE K1.S#='&X')
> > MINUS
> > (SELECT K2.P#
> > FROM SPJ K2
> > WHERE K2.J#=K.J#));
> >
> > 6.
> > SELECT DISTINCT K.J#
> > FROM SPJ K
> > WHERE NOT EXISTS ((SELECT K1.P#
> > FROM SPJ K1
> > WHERE K1.J#=K.J#)
> > MINUS
> > (SELECT K2.P#
> > FROM SPJ K2
> > WHERE K2.S#='&X'));
> >
> > 7.
> > SELECT DISTINCT K.J#
> > FROM SPJ K
> > WHERE NOT EXISTS ((SELECT K1.P#
> > FROM SPJ K1
> > WHERE K1.S#='&X')
> > MINUS
> > (SELECT K2.P#
> > FROM SPJ K2
> > WHERE K2.J#=K.J# AND K2.S#='&X'));
> >
> > 8.
> > SELECT DISTINCT K.J#
> > FROM SPJ K
> > WHERE NOT EXISTS ((SELECT K1.P#
> > FROM SPJ K1
> > WHERE K1.J#=K.J#)
> > MINUS
> > (SELECT K2.P#
> > FROM SPJ K2
> > WHERE K2.S#='&X' AND K2.J#=K.J#));
> >
> > 9.
> > SELECT DISTINCT K.J#
> > FROM SPJ K
> > WHERE NOT EXISTS ((SELECT S#
> > FROM SPJ K1
> > WHERE K1.J#=K.J#)
> > MINUS
> > (SELECT S#
> > FROM SPJ NATURAL JOIN P
> > WHERE COLOR='Red'));
> >
> > 10.
> > SELECT COUNT(DISTINCT J#) AS STEVILO
> > FROM SPJ
> > WHERE S#='&X';
> >
> > 11.
> > SELECT SUM(QTY) AS KOLICINA
> > FROM SPJ
> > WHERE P#='&X' AND S#='&Y';
> >
> > 12.
> > SELECT P#,J#,SUM(QTY) AS KOLICINA
> > FROM SPJ
> > GROUP BY P#, J#;
>
>
