On 5/7/07, [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 ......
>
While this is not a CakePHP-related question, first thing you should
do is make sure that Prototype is actually being loaded.
--
Chris Hartjes
My motto for 2007: "Just build it, damnit!"
@TheBallpark - http://www.littlehart.net/attheballpark
@TheKeyboard - http://www.littlehart.net/atthekeyboard
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