From: Subject: Google Groups: Prikaz niti "JMBG?" 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Prikaz niti "JMBG?" ------=_NextPart_001_00D3_01C1D0FE.A6028C60 Content-Type: application/octet-stream Content-Transfer-Encoding: quoted-printable Content-Location: http://groups.google.com/groups?hl=hr&frame=left&th=eb677d09cafd5296&prev=/groups%3Fq%3Djmbg%2Bgroup:hr.comp.*%26hl%3Dhr%26selm%3D9cgouf%2524k8c%25242%2540bagan.srce.hr%26rnum%3D6&seekm=9cgouf%24k8c%242%40bagan.srce.hr
3D""=20
=
Tema:
 JMBG?

News-grupa:
 hr.comp.programiranje.baze

3D""=20
3D""=20Natrag 3D""=20Bez okvir=E2 3D""=20Poredaj po datumu
3D""=20

3D""=20Ivica 27=20 Apr 2001

3D""=20|-Luka = Kladaric 27 Apr 2001
3D""=20|-Damir = Dzeko 27 Apr 2001
3D""=20| |-Ivica 27=20 Apr 2001
3D""=20| | \-Damir = Dzeko 27 Apr 2001
3D""=20| |   \-Tomislav = Vujec 28 Apr 2001
3D""| |     \-= Damir = Dzeko 29 Apr 2001
3D""=20| \-Robert = Petrunic 27 Apr 2001
3D""=20|   \-Damir = Dzeko 27 Apr 2001
3D""=20|     |-10 Robert = Petrunic 28 Apr 2001
3D""=20|     \-11 Tomislav = Vujec 28 Apr 2001
3D""=20|-12 Marko = Raspor 27 Apr 2001
3D""=20| \-13 Faruk = Smajevic 03 May 2001
3D""=20\-14 Maja = Martinjak 31 May 2001
3D""=20  |-15 Robert = Petrunic 01 Jun 2001
3D""=20  \-16 John = Smith 04 Jun 2001 =

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3D""=20
Poruke 11-16 iz=20 niti
Pre= thodih=20 10   
Sko=E8i na [ Po=E8= etak=20 niti ]

Message 11 in=20 thread
Od:Tomislav Vujec (tvujec@yahoo.com)
Tema:Re: = JMBG?=20
News grupe:hr.comp.programiranje.www
View this article=20 only
Datum:2001-04-28 16:14:02 = PST=20

Damir Dzeko =
<damir.dzeko@zg.hinet.hr> writes:

> No, najcesce je slucaj da se ljude uvodi u =
registre, maticne
> knjige ili stajaznam upravo u mjestu rodjenja.

Ne mjestu rodjenja, vec mjestu _upisa_. Ja imam sifru za vzd, a rodjen
sam u zgb. Isto tako i moj brat. Dok, ako mi treba izvod iz maticne
knjige rodjenih, vadim ga u Zagrebu.
Message 12 in=20 thread
Od:Marko Raspor (amal@amal.hr)
Tema:Re: JMBG? =
News=20 grupe:hr.comp.programiranje.baze, hr.comp.programiranje.www
View this article=20 only
Datum:2001-04-27 13:56:07 = PST=20
Ivica wrote:
>=20
> Pozdrav!
>=20
> Moze li mi netko napisati algoritam za provjeru JMBG-a?
> Naime, JMBG se sastoji od 13 znamenaka, a ona 13. je
> kontrolna (za slucaj da se krivo ukuca ostalih 12), a dobije
> se da se prethodne s ne=E8im mno=B5e, zbrajaju pa se od toga
> uzme neki modulo...
> Pa ako netko zna kako to ide, bio bih mu zahvalan.

http://josip.purger.com/software/jmbg/index.htm

Pozdrav,
--=20
Marko Raspor
AMAL d.o.o.
http://www.amal.hr/ - =
Want SuSE Linux?
"There's always a better way of doing something."
Message 13 in=20 thread
Od:Faruk Smajevic (fsmajevic@hotmail.com)
Tema:= Re:=20 JMBG?
News grupe:hr.comp.programiranje.baze, hr.comp.programiranje.www
View this article=20 only
Datum:2001-05-03 01:34:04 = PST=20
... i sajt je vrlo =
dobar,
Message 14 in=20 thread
Od:Maja Martinjak (maja.martinjak@= krapinske-toplice.hinet.hr)
Tema:Re:=20 JMBG?
News grupe:hr.comp.programiranje.baze, hr.comp.programiranje.www
View this article=20 only
Datum:2001-05-31 22:30:04 = PST=20
> Pozdrav !
>
> Moze li mi netko napisati algoritam za provjeru JMBG-a?
> Naime, JMBG se sastoji od 13 znamenaka, a ona 13. je
> kontrolna (za slucaj da se krivo ukuca ostalih 12), a dobije
> se da se prethodne s ne=E8im mno=B5e, zbrajaju pa se od toga
> uzme neki modulo...
> Pa ako netko zna kako to ide, bio bih mu zahvalan.



JMBG sastoji se od 13 znamenaka, 13. znamenka je konrolna.
Prvih 7 oznacava datum i godinu rodjenja u formatu ddmmggg a 13. koja je
kontrolna izracunava se na slijedeci nacin (znamenke su redom oznacene =
od z1
do z13):

z1*7+z2*6+z3*5+z4*4+z5*3+z6*2+z7*7+z8*6+z9+5+z10*4+z11*3+z12*2

dobiveni rezultat cjelobrojno se dijeli sa 11

ako je cjelobrojni ostatak dijeljenja veci od 1 onda se kontrolna =
znamenka
dobi kao rezultat oduzimanja cjelobrojnog ostatka od 11

ako je cjelobrojni ostatak =3D 1 kontrolna znamenka =3D 5
ako je cjelobrojni ostatak =3D 0 kontrolna znamenka =3D 0


Sto se tice programeskog dijela posla, to ovisi o programu u kojem se
programira.
Zasigurno postoji u Vasem progamu neka od funkcija koja daje cjelobrojni
ostatak djeljenja
a nakon toga stvar je jednostavna sa nekoliko if petlji.

Nadam se da ce Vam ovo pomoci ako Vam vec do sada nitko nije odgovorio.
Pozdrav !

M.M.
Message 15 in=20 thread
Od:Robert Petrunic (robert@petrunic.com)
Tema:Re: = JMBG?
News grupe:hr.comp.programiranje.baze, hr.comp.programiranje.www
View this article=20 only
Datum:2001-06-01 03:40:09 = PST=20
"Maja Martinjak" =
<maja.martinjak@krapinske-toplice.hinet.hr> wrote in
message http://groups.google.com/groups?selm=3D9f78da%2421g3%241%40=
as201.hinet.hr...
> ako je cjelobrojni ostatak =3D 1 kontrolna =
znamenka =3D 5
> ako je cjelobrojni ostatak =3D 0 kontrolna znamenka =3D 0

Nije istina - ako je cjelobrojni ostatak 1 onda je kontrolna znamenka 1 =
!!!!
Pogledaj moj proggy http://www.petrunic.com/JMBGer.html

Robert
Message 16 in=20 thread
Od:John Smith (js@ucsd.com)
Tema:Re: JMBG? =
News=20 grupe:hr.comp.programiranje.baze, hr.comp.programiranje.www
View this article=20 only
Datum:2001-06-04 05:30:45 = PST=20
kz =3D 11 - (suma MOD =
11)
(suma je dole dobro opisana)
IF kz =3D 10 THEN neispravan JMBG
IF kz =3D 11 THEN kz =3D 0
kontrolna znamenka trebala bi biti kz
HTH
JS


"Maja Martinjak" <maja.martinjak@krapinske-toplice.hinet.hr> wrote =
in
message http://groups.google.com/groups?selm=3D9f78da%2421g3%241%40=
as201.hinet.hr...
>
> > Pozdrav !
> >
> > Moze li mi netko napisati algoritam za provjeru JMBG-a?
> > Naime, JMBG se sastoji od 13 znamenaka, a ona 13. je
> > kontrolna (za slucaj da se krivo ukuca ostalih 12), a dobije
> > se da se prethodne s ne=E8im mno=B5e, zbrajaju pa se od toga
> > uzme neki modulo...
> > Pa ako netko zna kako to ide, bio bih mu zahvalan.
>
>
>
> JMBG sastoji se od 13 znamenaka, 13. znamenka je konrolna.
> Prvih 7 oznacava datum i godinu rodjenja u formatu ddmmggg a 13. =
koja je
> kontrolna izracunava se na slijedeci nacin (znamenke su redom =
oznacene od z1
> do z13):
>
> z1*7+z2*6+z3*5+z4*4+z5*3+z6*2+z7*7+z8*6+z9+5+z10*4+z11*3+z12*2
>
> dobiveni rezultat cjelobrojno se dijeli sa 11
>
> ako je cjelobrojni ostatak dijeljenja veci od 1 onda se kontrolna =
znamenka
> dobi kao rezultat oduzimanja cjelobrojnog ostatka od 11
>
> ako je cjelobrojni ostatak =3D 1 kontrolna znamenka =3D 5
> ako je cjelobrojni ostatak =3D 0 kontrolna znamenka =3D 0
>
>
> Sto se tice programeskog dijela posla, to ovisi o programu u kojem =
se
> programira.
> Zasigurno postoji u Vasem progamu neka od funkcija koja daje =
cjelobrojni
> ostatak djeljenja
> a nakon toga stvar je jednostavna sa nekoliko if petlji.
>
> Nadam se da ce Vam ovo pomoci ako Vam vec do sada nitko nije =
odgovorio.
> Pozdrav !
>
> M.M.
>
>
>
>
>
>
>
>

Pre= thodih=20 10   
Sko=E8i na [ Po=E8= etak=20 niti ]


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