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What variable corresponds to this if without woocommerce connection?
I want to use a script that was written for woocommerce taxonomies (not by me, I found it and used it many times), but this time when woocommerce is not connected. It turned out to replace taxonomies with others, but the console does not determine
var ajax_url = woocommerce_params.ajax_url;
for obvious reasons. I ran a search through the code, this variable is not manually defined anywhere, so it is out of the box.
And how to replace it for a regular wordpress without woocommerce? What does it correspond to? What record?
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In simple terms, this is a javascript variable that is needed to work with ajax.
It's in the admin by default.
Woocommerce adds it when installing the plugin.
That is, without Woocommerce, you need to add it yourself, since it is not included in your theme by default.
Anatoly gave you a link to the documentation above - everything is described in detail on this topic.
To simplify it further - you need to figure it out and add code like this -
// Подключаем локализацию в самом конце подключаемых к выводу скриптов, чтобы скрипт
// 'twentyfifteen-script', к которому мы подключаемся, точно был добавлен в очередь на вывод.
// Заметка: код можно вставить в любое место functions.php темы
add_action( 'wp_enqueue_scripts', 'myajax_data', 99 );
function myajax_data(){
// Первый параметр 'twentyfifteen-script' означает, что код будет прикреплен к скрипту с ID 'twentyfifteen-script'
// 'twentyfifteen-script' должен быть добавлен в очередь на вывод, иначе WP не поймет куда вставлять код локализации
// Заметка: обычно этот код нужно добавлять в functions.php в том месте где подключаются скрипты, после указанного скрипта
wp_localize_script( 'twentyfifteen-script', 'myajax',
array(
'url' => admin_url('admin-ajax.php')
)
);
}
This article will help you understand the principles of wordpress ajax, and also describes the nature of the ajaxurl variable and how to use it or replace it with your own.
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