If -(5) does not work and -5 does that should give you a clue as to
how to solve the problem.
You can put function calls anywhere that you can put a variable reference.
S.
On 10/04/06, Raffaele Viola <[EMAIL PROTECTED]> wrote:
> It doesn't work :-(
>
> Where can I put this besides a "Response assertion" ?
>
> -Raffo
>
> On 4/10/06, sebb <[EMAIL PROTECTED]> wrote:
> >
> > What does
> >
> > ${__intSum(${nb_aft},-(5),nb_aft2)}
> >
> > do?
> >
> > On 10/04/06, Raffaele Viola <[EMAIL PROTECTED]> wrote:
> > > I've tried,
> > >
> > > ${__intSum(${nb_aft},-5,nb_aft2)} works
> > > ${__intSum(${nb_aft},-(${nb_aft1}),nb_aft2)} doesn't works
> > >
> > > What can I do?
> > >
> > > Regards
> > > -Raffo
> > >
> > > On 4/10/06, sebb <[EMAIL PROTECTED]> wrote:
> > > >
> > > > Did that not work?
> > > >
> > > > I'd try something llike ${__intSum(23,-17,var)} first and then replace
> > > > the constants with variables.
> > > >
> > > > S.
> > > > On 10/04/06, Raffaele Viola <[EMAIL PROTECTED]> wrote:
> > > > > Ok therefore
> > > > > nb_aft =nb_be4 - nb_count
> > > > >
> > > > > becomes
> > > > >
> > > > > ${__intSum(${nb_be4},(-${nb_count}),nb_aft)} ?
> > > > >
> > > > > Thanks
> > > > > Raffo
> > > > >
> > > > > On 4/10/06, sebb <[EMAIL PROTECTED]> wrote:
> > > > > >
> > > > > > c=a-b=a+(-b)
> > > > > >
> > > > > > S.
> > > > > > On 10/04/06, Raffaele Viola <[EMAIL PROTECTED]> wrote:
> > > > > > > The quetions is also for the subtraction :
> > > > > > > How can I do c=a-b ?
> > > > > > >
> > > > > > > Thanks a lot
> > > > > > > -Raffo
> > > > > > >
> > > > > > > On 4/10/06, Raffaele Viola <[EMAIL PROTECTED]> wrote:
> > > > > > > >
> > > > > > > > Hi, all
> > > > > > > >
> > > > > > > > Can I, in my test plan, sum the value of 2 variable and puth
> > it in
> > > > a
> > > > > > 3rd?
> > > > > > > > For example:
> > > > > > > > a=2;
> > > > > > > > b=3;
> > > > > > > > c=a+b;
> > > > > > > >
> > > > > > > > Thanks, Raffo
> > > > > > > >
> > > > > > > >
> > > > > > >
> > > > > > >
> > > > > >
> > > > > >
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