我已經嘗試使用ajax調用我的PHP函數，但我不知道它的代碼錯誤是什么不起作用

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 --> 

以上是我的登錄表格。

以下是我的ajax電話

 //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; } }); } 

我不知道我的代碼是什么錯，即使代碼不能正常工作，它甚至沒有顯示表單空白輸入上的警報，單擊登錄按鈕后表單重新加載，請幫助我卡住非常糟糕。

Answer 1

嘗試下面的代碼，因為我剛從url中刪除了http：//。 希望這可以幫助。

//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: "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: "secret/signin.php",
                type : "POST",
                data: { sk:sk},
                success: function(r)
                {
                  if(r == '0')
                  {
                     window.location.href = "main/index.php";
                  }
                  else
                  {
                    window.location.href = "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;
        }


      });
    }

Answer 2

Ajax調用中的Type屬性定義了兩次。

請使用調試工具（如Firebug）來調試xhr請求，以了解它們是否被發送。 您還可以查看可能提示錯誤的請求的響應。

Answer 3

在我將javascript文件的序列更改為我的表單的html文件后，代碼正在運行。

我把我的javascript代碼文件放在后面

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

它工作得很好。