How to post the viewmodel to action method using JQUERY AJAX in MVC

Question

I want to achieve create (INSERT) screen using $.POST or $.AJAX.

Note: code is working fine without AJAX call..it is already there.. now i need to do ajax call and s

Answer 1

If the id of your form is frm, then it should be referenced as follows

data:   $("#frm").serialize(),

Answer 2

For id Selector you have to use "#" with id.

This will work :

$.ajax(
  {
      url: '@Url.Action("CreateProduct","ProductManagement")',
      dataType: 'json',
      contentType: 'application/json; charset=utf-8',
      type: 'post',
      data:   $('this').serialize(), OR  $('#frm').serialize(),   <-----------------
      success: function () { alert('s'); },
      error: function (e1, e2, e3) { alert(e2); alert(e1 + ' ' + e2 + ' ' + e3); alert('failure event'); }
  }
 );

OR Try this :

var form = $('#frm');
$.ajax({
 cache: false,
 async: true,
 dataType: 'json',
 contentType: 'application/json; charset=utf-8',
 type: "POST",
 url: form.attr('action'),
 data: form.serialize(),
 success: function (data) {
 alert(data);
   }
 });

Answer 3

You have forgot to add id selecter:

$.ajax(
      {
          url: '@Url.Action("CreateProduct","ProductManagement")',
          dataType: 'json',
          contentType: 'application/json; charset=utf-8',
          type: 'post',
          data:   $('#frm').serialize(),                      // $('#frm');
          success: function () { alert('s'); },
          error: function (e1, e2, e3) { alert(e2); alert(e1 + ' ' + e2 + ' ' + e3); alert('failure event'); }
      }
    );