I have "add to favorite" feature in my wordpress project, I am using <form> to submit the feature. Each time I click on add to fav button it works, but the page is always reloaded which is annoying so I decided to use AJAX, but I can not connect it with my PHP file.
This is PHP file, where I wrote function for the feature
if (! function_exists('favorite_button_icon')) {
function favorite_button_icon() {
echo apply_filters( 'favorite_button_icon', get_the_favorite_button_icon() );
}
}
if (! function_exists('get_the_favorite_button_icon')) {
function get_the_favorite_button_icon() {
$output = '';
$action = $currentPath; // window.location.href
$name = is_favorite() ? 'remove_favorite' : 'add_favorite';
$class = 'favorite-button-pro' . ( is_favorite() ? ' favorite' : '' );
$icon = is_favorite() ? 'Added' : 'Add';
$output .= '<form method="post" id="favorite_user_post"
class="favorite_user_post" action="'. $action .'">';
$output .= '<button id="submit-favorite" type="submit" name="'. $name .'"
value="'. get_the_ID() .'" class="'. $class .'">';
$output .= $icon;
$output .= '</button></form>';
return apply_filters( 'get_the_favorite_button_icon', $output );
}
}
And I am just calling that function in html
<div class="add-to-favorites-option" >
<?php favorite_button_icon() ?>
</div>
Everything above works, but as I said, page is being reloaded.
This is my attempt to use AJAX
jQuery('#favorite_user_post').submit(function (e) {
e.preventDefault();
var form = jQuery(this);
var url = form.attr('action');
jQuery.ajax({
type: 'POST',
url: url,
data: form.serialize(),
success: function (data) {
console.log(data);
},
error: function () {
console.log('Fail');
},
});
});
I get absolutely no fire on submit, am I missing something?
You need to use the wp_ajax_ hook: https://developer.wordpress.org/reference/hooks/wp_ajax_action/
Simple example below:
You PHP code to process the ajax request:
add_action( 'wp_ajax_my_action', 'my_action' );
function my_action() {
// code that captures your ajax POST request
wp_die(); // this is required to terminate immediately and return a proper response
}
Your Ajax call:
jQuery('#favorite_user_post').submit(function (e) {
e.preventDefault();
var form = jQuery(this);
// note the added "my_action" to tell the server what function to fire (my be a better way to append this to your form data)
var url = form.attr('action') + "&=my_action";
jQuery.ajax({
type: 'POST',
url: url,
data: form.serialize(),
success: function (data) {
console.log(data);
},
error: function () {
console.log('Fail');
},
});
});