sorry maybe I did explain it not good enough.
The problem is that if a user has entered data in form 2 goes back to form 1
and forward to form 2 again, the data he entered once in form 2 is lost. And
I do not find a way how the get the data out of form 2 because the back
button is placed in form 3 (same site as form2 ) to allow a different action
since we have to link to another site then in form 2.
sounds confusing? I know but I hope I explained it ok
Andy
----- Original Message -----
From: "Alberto Serra" <[EMAIL PROTECTED]>
Newsgroups: php.general
Cc: <[EMAIL PROTECTED]>
Sent: Sunday, July 07, 2002 1:21 PM
Subject: Re: [PHP] transporting variable via post to another site
> Привет!
>
> andy wrote:
> > Everything works fine exept of error handling. Which means if a user
wants
> > to go back from step 2 to one and has already filled in some data in
site 2
> > he will loose this data for sure. It is not possible to transport the
data
> > via get anymor because the text is way to long.
>
> Okay, this means Form 1 can be entered from two directions:
> 1) the usual sequence (that is, the way users usually get to it)
> 2) getting back from Form 2
>
> If you place a button on Form2 saying "go back to form 1" when your user
> presses it just post your current data back to the script of form1.
> *PLUS* you add a hidden field called *FLAGBACK* (or whatever you want to
> call it like.
>
> when the script that prepares form1 is called it must check for
> *FLAGBACK* being present. If not, it will do the same old stuff, if yes
> you just write a new branch that will put the data back into form 1. Easy.
>
> *BUT* all this will never protect your user in case he just presses
> *back* on his browser. Make sure this is well explained on the page.
>
> пока
> Альберто
> Киев
>
>
> --
>
>
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