On Wednesday, April 24, 2002, at 02:59 PM, Julio Nobrega Trabalhando
wrote:
> I have a simple function that returns a message:
>
> function showError ($mensagem)
> {
> return '<div class="erro">' . $mensagem . '</div>';
> }
>
> And this is how I use it:
>
> if (isset($_SESSION['forum']['error']['insert'])) {
> echo showError($_SESSION['forum']['error']['insert']));
> }
>
> But I want that when the function is used the session variable is
> destroyed. A simple unset($mensagem) inside showError() gives me back
> this:
>
> Warning: Undefined variable: mensagem in
> c:\www\lib\sistema\formularios\formularios.inc.php on line 92
The problem is that there is no variable called $mensagem in your
function. You are using a reference to something else with the name
$mensagem, but you are not passing the value by reference -- try this:
function showError(&$mensagem)
// note the '&' before $mensagem
{
return '<div class="erro">' . $mensagem . '</div>';
}
And let me know if that works or doesn't work. (It's kind of a hunch,
hard to tell from your code if it will work.)
Erik
----
Erik Price
Web Developer Temp
Media Lab, H.H. Brown
[EMAIL PROTECTED]
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