“Ajaxcall” Kode Jawaban

Ajaxcall

//ajaxcall function 
var $ajaxCall = function (data, url, success, error) {
  $.ajax({
    type: "POST",
    async: true,
    url: url,
    data: data,
    success: success,
    error: error
  });
};

//usecase
$ajaxCall({ data: data }, "/pass/url/here", function (response) {}, function (error) {});