I have a select, lets call it A. When i choose an option from A, with ajax, i load an another select with options into a div. Lets call this second select B.
I have some ajax onchange code on B, but it does nothing, and it gives no error also.
I think, the "site doesnt see the B select", because it isnt in the source code, ajax loads it into a div.
What would be the solution for this?
$('#fizetes_select').on('change', function() {
    var SelectedValue = this.value;
    if(SelectedValue != 0 )
    {
        $.ajax({
             type: 'POST',
             url: 'files/get_fizetes_text.php',
             data: { id: SelectedValue },
             dataType: "html",
             cache: false,
             beforeSend: function(){
                $('#preloader_fizetes').show();
             },
             success: function(data)
             {
                 var result = $.trim(data);
                 $('#fizetes_result').html(result);
             },
             complete: function(){
                $('#preloader_fizetes').hide();
             }
        });
    }
    return false;
});
Thats the onchange code for the B select, i only want to display some text with it. The ID-s are correct, i didnt write it bad, i chekced. Its in document ready.
 
    