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Getting illegal invocation, after using $.ajax inside of $(doocument).keypress function


By : Louis
Date : October 18 2020, 06:10 PM
I wish did fix the issue. illegal invocation is usually caused when you pass a non plain object to $.ajax data parameter.
window.location is not a string rather a Location object, you should use window.location.href instead.
code :


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Ajax - illegal invocation error


By : user103474
Date : March 29 2020, 07:55 AM
may help you . There's no standard serialisation using x-www-url-form-encoded for an array of strings.
Either use some additional serialisation (e.g. JSON) to convert the array into a string and then de

Illegal Invocation in AJAX call


By : user3448560
Date : March 29 2020, 07:55 AM
wish helps you I am building a web-app in JQuery that needs to communicate with a server. I have used this exact same code several times, the only thing that's changed is the data and the code performed on success. , See this is causing the issue:
code :
var name = $("#ncName"); // you have to get the value/text of it
addCourse(name);
var name = $("#ncName").val(); // $("#ncName").text(); in case if it is not a form control.
addCourse(name);

Illegal invocation with $.get inside $.ajax success


By : Someone_from_unknown
Date : March 29 2020, 07:55 AM
wish helps you The only multi-argument syntax for $.get is jQuery.get( url [, data ] [, success ] [, dataType ] )
where data can be a simple object, however you can't send a DOM element, or a jQuery object
code :
$.get($('#articles').data('url'), {
  after: $('#articles')
});
$.get($('#articles').data('url'), function( data ) {
    $('#articles').after( data )
});

Ajax error `Uncaught TypeError: Illegal invocation when making ajax call


By : sanjay chouhan
Date : March 29 2020, 07:55 AM
To fix the issue you can do I am using the script below to process my login form and send login request by ajax but I get the error. I have never experienced this error before. The code is here below: , You need to set
code :
function loginReq() {
  var username = $("#username").val();
  var password = $("#password").val();
  // Checking for blank fields.
  if( username =='' || password ==''){
      $('input[type="text"],input[type="password"]').css("border","2px solid red");
      $('input[type="text"],input[type="password"]').css("box-shadow","0 0 3px red");
      $('#login_alert').show();
      $("#login_alert").fadeTo(2000, 500).slideUp(500, function(){
          $("#login_alert").slideUp(500);
      });
  } else {
      $.ajaxSetup({
          headers: {
              'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
          }
      });
      var link="login";
      var formData = new FormData(document.getElementById('#login_form'));
      $.ajax({
          type: 'post',
          dataType: 'html',
          url: link,
          cache: false,
          data: formData,
          processData: false,
          //^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
          success: function (result) {
              var obj = jQuery.parseJSON(result);
              var status = obj['success'];
              if(status == 'success'){
                  window.location.href = "{{ url('dashboard') }}";
              }else{
                  $("#login_alert2" ).show();
                  $("#login_alert2").fadeTo(2000, 500).slideUp(500, function(){
                      $("#login_alert2").slideUp(500);
                  });
              }
          }
      });
  }
}

$('#btn').on('click', function(e) {
  loginReq();
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<form id="login_form">
    User name: <input type="text" id="username" value="username">
    Password: <input type="password" id="password" value="pasword">
    <button type="button" id="btn">Send Ajax</button>
</form>

PHP and Ajax: Illegal invocation


By : user3108410
Date : March 29 2020, 07:55 AM
I hope this helps . You're passing a file to data and jQuery is erroring while attempting to encode it as form data.
You should:
code :
var form = $('#uploadform')[0];
var form_data = new FormData(form);

$.ajax({
    url : "upload.php",
    type: "POST",
    data: form_data,
    processData: false,
    contentType: false,
    success: function(response) {
        $('#success_message ').text(response);
    }
});
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