基于条件的Zend表单元素

Question

I have came across with one problem i need to show different form elements

• Based on different condition to the user i know it will be easy in core php but want to do in zend environment

EXAMPLE:

If person is disabled we will show two radio buttons

$disable = new Shaadi_Form_Element_radio('disablitiy');
$disableArr = array(""=>" Doesn't Matter","Y"=>" show disabled");

IF he is not disabled

$disable = new Shaadi_Form_Element_radio('disablitiy');
$disableArr = array(""=>" Doesn't Matter","N"=>" do not show disabled");

I want this code to be done in form how can I optimize this please help me

Answer 1

Whenever a form - or any object, for that matter - is dependent upon some information external to the form, I usually pass that information in the form's constructor. Then I inspect it later in init() when I am building the form.

Example:

class My_Form extends Zend_Form
{
    protected $hasDisability;

    public function __construct($hasDisability)
    {
        $this->hasDisability = (bool) $hasDisability;
        parent::__construct();
    }

    public function init()
    {
        // Add all your other elements
        // Blah, blah

        // Add the element that is dependent upon the $hasDisability value
        $disable = new Shaadi_Form_Element_radio('disablitiy');
        $disableArr = $this->hasDisability
            ? array(""=>" Doesn't Matter","Y"=>" show disabled")
            : array(""=>" Doesn't Matter","N"=>" do not show disabled");

        // Add the $disableArr into the radio element
        $disable->setMultiOptions($disableArr); 
    }
}

Usage - perhaps in a controller - is then something like:

$form = new My_Form(true);  // for a disabled used
$form = new My_Form(false);  // for a non-disabled user