Show modal/popup on a certain date?

Support
1

Hello,
is it possible to show a modal or popup on a certain date?
I want to show a message on the Dutch holiday’s date’s
I tried mxcalendar and play with resource Publish Date function but no success

Stefan

2

Yep should be possible. You’ll need to create a snippet with something like this in it (untested) and a chunk for your modal.

$today = date("d/m/Y"); 
$holiday = '05/12/2017';

if ($today === $holiday) {
   	$output = $modx->getChunk('modal_chunk');
   	return $output;
}

Then just call the snippet on the page or template you want to display it on. If you want multiple holidays then you could transform the snippet above into a switch case.

3

@lkfranklin
Thanks your (and my first) snippet works great. And i will find out how the “switch case” works

Stil have a question. How to trigger the modal?

The snippet calls the chunk(that work fine) but how to trigger the modal?

4

Triggering the modal is not related to modx specifically.

You can use whatever client-side solution that best fits your needs.

5

If the chunk only renders if the date matches then you could have it as display:block in your css by default, or maybe add a class to force display: block.

If you want it to be triggered by a button then you could give the modal an id and then use JS to show/hide the modal.

I just uploaded a video yesterday where I did something similar, check it out here at this time stamp:

6

Thanks for the help. Its works perfect!!!
Edit:

If someone wants to help me to transform the snippet to a switch case for multiple holiday’s?
I am grateful for that
Stefan

7

Please see: https://www.w3schools.com/php/php_switch.asp

So something like this I think:

switch ($date) {
    case "01-01-2019":
        // render chunk
        break;

    case "01-02-2019":
        // render chunk
        break;
};

Maybe an array is easier?

$today = date("d/m/Y"); 
$holidays = array(
	"01-01-2019" => 'Holiday name here',
	"01-02-2019" => 'Holiday name here',
	"01-03-2019" => 'Holiday name here'
);

if( array_key_exists( $today, $holidays ) ){
 // render chunk
}

(code = untested)