No Nate I jus havit inside the conditions of the while loop thats why iam
worried since shoulndt do that what do you suggest problem with mysql???
----- Original Message -----
From: Nate LaClaire <nate@m...>
To: professional php <pro_php@p...>
Sent: Jueves, 22 de Marzo de 2001 12:35 p.m.
Subject: [pro_php] RE: to all and to Adam Lang problem with a


> Jorge,
>
> The only thing that I can think of that would cause this problem is having
> "$result=mysql_fetch array($query);" somewhere above your while loop. This
> would cause the first record to be placed into $result and then be
replaced
> by the second record when your while loop is invoked. You don't have a
line
> like that above the while loop, do you?
>
> Nate LaClaire
>
> ---------------------------------
> Netwalker Internet Services
> http://www.netwalkerinternet.com/
> ---------------------------------
>
>
> -----Original Message-----
> From: Info [mailto:info@d...]
> Sent: Thursday, March 22, 2001 11:17 AM
> To: nnetwalker@h...
> Subject: [pro_php] to all and to Adam Lang problem with a
>
>
> I solve the problem like this but It seems sopmething wrong
> i wa using a while loop
> like this
> while($result=mysql_fetch array($query)){
> print all the results.....an the first result of the set was missing
> }
> now I am doing a do while loop
> do{
> print the results}while($result=mysql_fetch array($query));this prints
> everything fine;
> why is this hapening i now that the do while loop get executed at least
one
> time but why the while loop doesnt show the first result or the query???
It
> mus show it right??
>
> ----- Original Message -----
> From: Jorge Cordero <jorge@d...>
> To: professional php <pro_php@p...>
> Sent: Miércoles, 21 de Marzo de 2001 04:36 p.m.
> Subject: [pro_php] when querying a data base mysql_fecth_array jumps the
1st
> record???
>
>
> > I ha ve a database  with cd music albums
> > author ,title, group, cd_number, and song_num also there is an auto
> > increment id_num column.
> > if i query the data base through my web page on the cd_number the first
> song
> > of  the cd never shows.
> > but when query by any of the other fields everything is perfect.
> > if i query directly on the server through telnet all of them are fine
> > my query is so simple that cant see why this is happening
> >
> > $query=mysql_query("SELECT * FROM tablename WHERE cd_num=$cd_num order
by
> > song_num");
> > this is the same query I am using for everything else the only change is
> the
> > WHERE clause.
> >
> > to extract the result I am using
> >
> > while($result=mysql_fetch_array($query)){
> > results here (html table)
> > }
> > This have never happened before so if some out there knows whats going
on
> >
> > Thanks for your help
> >
> > Jorge
> >
> > ----- Original Message -----
> > From: Adam Lang <aalang@r...>
> > To: professional php <pro_php@p...>
> > Sent: Viernes, 16 de Marzo de 2001 08:39 a.m.
> > Subject: [pro_php] Re: Variables PHP <=> JavaScript
> >
> >
> > > You need to have a hidden input type that your PHP will increment each
> > time
> > > the page is recalled.  Then, have an if or case statement that checks
> the
> > > value of that hidden type to take appropriate action.
> > >
> > > Adam Lang
> > > Systems Engineer
> > > Rutgers Casualty Insurance Company
> > > http://www.rutgersinsurance.com
> > > ----- Original Message -----
> > > From: "Thor Vadstein" <thor@l...>
> > > To: "professional php" <pro_php@p...>
> > > Sent: Thursday, March 15, 2001 10:21 PM
> > > Subject: [pro_php] Re: Variables PHP <=> JavaScript
> > >
> > >
> > > > > not directly no.  Remember that Javascript is on the client and
PHP
> is
> > > > on
> > > > > the server.  Thus Javascript can/will make either a POST or a GET
to
> > the
> > > > > server passing variables the appropriate ways (querystring, etc..)
> > from
> > > > > there you can use them normally in your PHP code.
> > > > >
> > > >
> > > > I will have to call the page again with the variables as part of the
> > url,
> > > > is that what you mean?
> > > > Is there an easy way to separate between first and second
(subsequent)
> > > > calls to a page?
> > > >
> > > > Regards,
> > > > ThorV.
> >
> >
>