Maybe php is interpreting $F as a variable, so try using single quotes
for the code block, like echo $javascript->codeBlock(' ...
$F("sectionName") ...
On May 7, 10:06 am, "[EMAIL PROTECTED]"
<[EMAIL PROTECTED]> wrote:
> Hi
> I have this view:
>
> <?php
> echo $javascript->codeBlock("
> function createNewArticle(id) {
> var name = $F('sectionName');
> alert(name);
> }
> ");
> ?>
>
> <input type="text" id="sectionName" size="25">
>
> When I try this view with my browser I get this error:
>
> Notice: Undefined variable: F in ......
>
> How I can resolve this ?
>
> Many Thanks
> Marco
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