i have used
$need = param('dif') - $score.
i didin't know i could use params in a calc like that.
problem solved, cheers
On Monday 25 Mar 2002 6:38 pm, Dave Storrs wrote:
> Hi Matthew,
>
> First of all, subtracting scalars is perfectly valid, so the actual error
> must be something else. He
Hi Matthew,
First of all, subtracting scalars is perfectly valid, so the actual error
must be something else. Here are three things you should check.
1) First, you don't have a semicolon at the end of that line.
2) Second, if you are operating under 'use strict', you will need to
predeclare yo
Matthew Harrison wrote:
>
> I have two scalars created from passed params. i want to create a third
> variable by subtracting the two scalars. how can i do this? i have tried:
>
> $need = $dif - $score
>
> but i get a 500 server error and my error log says that i cannot use the
> '-' operator o
I have two scalars created from passed params. i want to create a third
variable by subtracting the two scalars. how can i do this? i have tried:
$need = $dif - $score
but i get a 500 server error and my error log says that i cannot use the
'-' operator on a scalar variable. can anyone suggest
