Adding inline JavaScript to form input element in CakePHP

Question

I'm trying to do this:

echo $this->Form->input("filter.wijk", array(
    'onChange' => 'this.form.submit()', 
    'type' => 'select', 
    'multiple' => 'checkbox', 
    'options' => $wijkOpties, 
    'label' => false)
);

However, the onChange attribute does not appear in the resulting HTML. The documentation for FormHelper::input tells us this:

* ### Options
*
* See each field type method for more information. Any options that are part of
* $attributes or $options for the different **type** methods can be included in `$options` for input().i
* Additionally, any unknown keys that are not in the list below, or part of the selected type's options
* will be treated as a regular html attribute for the generated input.

Am I interpreting that last sentence in the wrong way?

Well I don't believe it's satisfied with the 'multiple' => 'checkbox', your code is just fine, it's simply that when the list is changed to multiple checkboxes list the onChange is not copied too ! Try 'multiple' => true, and your code will work fine just as stated in the documentation. — Ayman Bedair
– Ayman Bedair, Commented Oct 21, 2014 at 8:59
Thanks, that does work as far as the JavaScript is concerned, but now they are not checkboxes anymore and I really need checkboxes here. — Jasper Driessens
– Jasper Driessens, Commented Oct 21, 2014 at 9:14

Ayman Bedair · Accepted Answer · 2014-10-21 09:40:25Z

1

You have 2 options:

1- Give up on the checkboxes and simply get a normal select list element/box:

echo $this->Form->input("filter.wijk", array(
                        'onChange' => 'this.form.submit()', 
                        'type' => 'select', 
                        'multiple' => true, 
                        'options' => $wijkOpties, 
                        'label' => false)
);

2- Do a workaround for a valid checkboxes list (Something like):

$i = 0;
foreach($wijkOpties as $value => $option){
    echo $this->Form->input('filter.wijk.' . $i++, array(
                                'onChange' => 'this.form.submit()',
                                'label' => $option,
                                'value' => $value,
                                'type' => 'checkbox',
    ));
}

`

answered Oct 21, 2014 at 9:40

Ayman Bedair

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