Yes the problem is exactly that php is interpreting $F as a variable
On 7 Mag, 16:06, bernardo <[EMAIL PROTECTED]> wrote:
> 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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