我已经尝试使用ajax调用我的PHP函数，但我不知道它的代码错误是什么不起作用

Question

 !-- Main Page Starts Here -->
  <section class="container">
    <div class="row">
      <div class="centerlogin">
         <div class="frmlogin">
          <form role="form" name="signin" id="signin" method="post" action="#">
          <div class="headtab"><h3>Login</h3></div>
          <ul>    
          <li><i class="glyphicon glyphicon-user"></i>&nbsp;&nbsp;&nbsp;&nbsp;<input type="text" id="email" name="username" class="usern" placeholder="Enter Username"></li>
          <li><i class="glyphicon glyphicon-lock">&nbsp;</i><input type="password" id="pwd" name="password" class="passn" placeholder="Enter Password"></li>
          <li><button class="subn" id="btnSubmit">Login</button></li>
         </ul>
         </form>
        </div>

      </div>

    </div>
  </section>
  <!-- Main Page Ends Here --> 

the above is my login form.

and below is my ajax call

//ajax calls start below
$(document).ready(function () {
        $("#btnSubmit").click(function (e) {
          e.preventDefault();
        var email = $("#email").val();
        var password = $("#pwd").val();
        var pwd = $.md5(password);
        auth(email, pwd);
        });
        });

        //authenticate function to make ajax call
        function auth(email, pwd) {
        $.ajax
        ({
        type: "POST",
        url: "https://localhost/main/web/sign-in",
        dataType: 'json',
        type : "POST",
        data: { email: email,pwd: pwd },
        success: function (r) {
          //console.log(r);
          if(r.status == '0')
          {
            var sk=r.sk;
            $.ajax({
                type: "POST",
                url: "http://localhost/main/secret/signin.php",
                type : "POST",
                data: { sk:sk},
                success: function(r)
                {
                  if(r == '0')
                  {
                     window.location.href = "http://localhost/main/index.php";
                  }
                  else
                  {
                    window.location.href = "http://localhost/main/login.php";
                    alert('Something Went Wrong.Please Try Again!');
                  }
                }

              });
          }
          else if(r.status == '401')
          {
            alert("Incorrect Email/Password");
            $("#signin")[0].reset();
          }
          else
          {
            alert("User Doesn't exist");
            $("#signin")[0].reset();
          }
        return false;
        }

        
      });
    } 

I dont know whats wrong with my code even the code its not working, it is not even showing the alerts on form blank inputs and form gets reload after clicking login button,Please help me stuck very badly.

</div>

Answer 1

Type property in Ajax call is defined twice.

Please use debugging tool, such as Firebug to debug xhr requests to understand if they are being sent or not. You can also view the responses of those requests which might hint errors.