Question

我使用下面的脚本来处理我的登录表单并通过ajax发送登录请求，但是我收到错误。我以前从未经历过这个错误。代码如下：

Uncaught TypeError: Illegal invocation
    at e (jquery.min.js:4)
    at Ab (jquery.min.js:4)
    at Function.r.param (jquery.min.js:4)
    at Function.ajax (jquery.min.js:4)
    at loginReq (admin:95)
    at HTMLButtonElement.onclick (admin:15

2）

<script language="javascript" type="text/javascript">

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,
              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);
                    });
                }
              }
        });
  }
}
</script>

有人可以帮助我解决问题！

Answer 1

您需要设置

processData 为false：默认情况下，作为对象传入数据选项的数据（技术上，不是字符串）将被处理并转换为查询字符串，适合默认内容类型＆＃34; application / x-www-form-urlencoded＆＃34;。如果要发送DOMDocument或其他未处理的数据，请将此选项设置为false。

＆＃13;

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();
});

＆＃13;

<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>

＆＃13;

Ajax错误`Uncaught TypeError：进行ajax调用时非法调用

1 个答案: