我需要如何将其转换为普通的 javascript 或 jquery？

Question

I'm new to JavaScript and jQuery. I'm currenly having the following code in my javascript file, however it doesn't seem to be working. I'm using this from prototype.js :

        var url = '/sip/TnsViewScreenResponse';
        var myAjax = new Ajax.Request(url, {
                method: "post",
                headers:{
                    'X-Requested-By': 'XMLHttpRequest'
                 },
                 parameters: "tin=" + tin,                 
                success: function transResult(response) {
         document.getElementById('tinVersionsOf_' + tin).innerHTML 
         = response.responseText;                    
         document.getElementById('ajax_loading_img').style.display 
         = 'none';
         document.getElementById('tinVersionsOf_' + tin).style.display = 
         'block';
         },
         error: function transResult(response) {
         document.getElementById('ajax_loading_img').style.display = 'none';
         alert('Failure: Problem in fetching the Data');
               },              
            }
         );         
         return false;

This seems to be conflicting with the other jQuery files being used in the file, so I want to convert this to plain JavaScript or jQuery. I have tried the below but it doesn't seem to be working. How can I make this right ?

var url = '/sip/TnsViewScreenResponse';
var myAjax = $.ajax({
  type: "POST",
  url: url,
  data: tin,
  success: function transResult(response) {
         $('#tinVersionsOf_' + tin).html(response.responseText);                    
       $('ajax_loading_img').css("display","none") ;
       $('#tinVersionsOf_' + tin).css("display","block");
         },
   error: function transResult(response) {
         $('#ajax_loading_img').hide();
         alert('Failure: Problem in fetching the Data');
               },              
            }
});

The above code is getting skipped while being parsed in the browser, which I had checked with inspect element option in Google chrome.

Answer 1

Try This

 $.ajax({ type: "POST", dataType: "json", contentType: "application/json", url: "/sip/TnsViewScreenResponse", data: JSON.stringify({ mydata: tin }),//where tin is ur data success: function (result) { //include your stuff }, error:function(error) { // include your stuff } });

 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>