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- // Самый простой код, работает с JQuery 3.2.1
- <div id="mybutt">Жми!<br></div>
- <div id="result_form"></div>
- <script>
- mybutt.onclick = function() {
- console.log('клик');
- var aa = sendAjaxForm('result', 'contact', "http://my.loadbot.ru/aj/test.php");
- console.log(aa);
- }
- function sendAjaxForm(result_form, ajax_form, url) {
- $.ajax({
- url: url,
- type: "POST",
- data: "данные",
- success: function(data){
- console.log(data)
- },
- dataType: 'html'
- });
- }
- </script>
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