Yes this is true.
But anyway , I manage to solve the pb by using in the view sthg like :
<input name="id" type="hidden" value={{=request.args[0]}}>
But I decided to leave this solution because it seems that it overrides some
issue even if it worked for me. So now I'm using {{form.custo.begin}}.....
because a lot of good is done by the framework that I do not want to do
myself.
JmiXIIII
2010/4/23 Paul Wray <paul.w...@det.nsw.edu.au>
> I too have tried this without success.
>
> I seems that you cannot use a custom form together with SQLFORM with
> record=
> Is that true?
>
> Paul
>
>
> On Apr 9, 6:14 am, JmiXIII <sylvn.p...@gmail.com> wrote:
> > Hello,
> >
> > I'm using a SQLFORM with a html custom (as described in book/7.2 =>
> > SQLFORM in HTML)
> > My fonction inside my controller is :
> >
> > def ncmodif():
> > ### Formulaire de saisie des NC de production
> > rec=db(db.NC.id==269).select()[0]
> > form=SQLFORM(db.NC,record=rec)
> > if form.accepts(request.vars,formname='test'):
> > response.flash = 'form accepted'
> > elif form.errors:
> > response.flash = 'form has errors'
> > else:
> > response.flash = 'please fill the form'
> > return dict()
> >
> > I have hardcoded db.NC.if==269 for the moment to see how it works.
> >
> > My view is something like:
> > <form><input name="FieldName"/>....<input type="hidden"
> > name="_formname" value="test" />....
> >
> > My question is there a simple way to make this form work as it works
> > with the standard way (ie when using {{=form}}) to update the record
> > via the custom form ?
> > Can I have the input field prepopulated and updated when I submit
> > since ncmodif() does not return any value ?
> >
> > Thanks
> >
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