Re: Quickform: image as submit button?

On 12/14/06, Vignesh M P N <manivig@iit.edu> wrote:

Hi ========= Question 1 ========= I need to use image as submit button in my Quickform, so I assigned the "value" attribute as "submit". Please find below the way I define the image element. $loginimgargs = array('value' => 'submit', 'id' => 'loginimg',

    'onmouseover' =>

'FP_swapImg(1,0,/*id*/\'loginimg\',/*url*/\'img/button15.gif\')',

    'onmouseout' =>

'FP_swapImg(0,0,/*id*/\'loginimg\',/*url*/\'img/button14.gif\')',

    'onmousedown' =>

'FP_swapImg(1,0,/*id*/\'loginimg\',/*url*/\'img/button16.gif\')',

    'onmouseup' =>

'FP_swapImg(0,0,/*id*/\'loginimg\',/*url*/\'img/button15.gif\')',

    'fp-style' => 'fp-btn: Embossed Capsule 1; fp-font-style: Bold;

fp-transparent: 1; fp-proportional: 0',

    'fp-title' => 'Log in');

$formAuth->addElement('image', 'loginimg', 'img/button14.gif', $loginimgargs); But when I submit the form by clicking the "loginimg", isset($_POST['loginimg']) returns false! I tried printing all the POST variables of the form and it printed 2 variables "loginimg_x" and "loginimg_y" whose values are the respective coordinates. Now I can simply use "isset($_POST['loginimg_x'])" for the code to work fine, but is it right way to do it or am I doing anything wrong here?

Try $form->exportValue('loginimg');

========= Question 2 ========= I validate the Quickform using the rules. But I have two submit buttons in the form, for example, "Login" and "New account" buttons. In this case, the rules need to validated only when I click "Login" button and the rules shouldn't be validated when I click "New account" button. How to achieve this?

I suggest checking which button was clicked before adding the rules (which in turn needs to be before validate() is called). -- Justin Patrin