通过AJAX重定向到PHP中的另一个页面

流黑手党

我的页面正在从javascript收集信息，然后将其发送到PHP，然后再发送到MySQL，问题是我希望根据数据库中的数据将其重定向到不同的页面，我尝试使用标头，但这只是向我展示警报中另一页的整个HTML代码，我不想要那样。我希望它根据情况重定向到一页或另一页

HTML（Login.html）

<div class="wrap-input100 validate-input" data-validate = "Enter username">
<input class="input100" type="text" id="user" name="username" placeholder="Email">
<span class="focus-input100" data-placeholder="&#xf207;"></span>
</div>

<div class="wrap-input100 validate-input" data-validate="Enter password">
<input class="input100" type="password" id="pass" name="pass" placeholder="Password">
<span class="focus-input100" data-placeholder="&#xf191;"></span>
</div>

<div class="container-login100-form-btn">
<a class="login100-form-btn" id = "logBtn">
Login
</a>
</div>

脚本

$('#logBtn').click(function(event){ 
    user = document.getElementById("user").value;
    password = document.getElementById("pass").value;

    $.ajax({
        type:"POST",
        url:"login.php",
        async: false,
        data: {user:user,password:password},
        success: function(data){
        alert(data);
        //window.location = '../Main/index.html';
        }
        });
    });

的PHP

<?php
$servername = "localhost";
$username = "root";
$password = "tbjdjkdl";
$dbname = "dbbbbbb";

$conn = new mysqli($servername, $username, $password, $dbname);

$user = $_POST['user'];
$pass = $_POST['password'];

$sql = "SELECT * FROM users WHERE email='$user' AND clave='$pass'"; 
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    $sql = "SELECT * FROM users WHERE email='$user' AND clave='$pass' AND permisos='Administrador'";
    if (mysqli_num_rows($result) > 0){
        echo "admin";
        header('Location: ../Main/index.html');
        exit;
    }
    else{
    echo "user";
    header('Location: ../Main/startemp.html');
    exit;
    }

 } else {
    $msg = "username/password invalid";
    echo $msg;
 }

mysqli_close($conn);
?>

中谷根

使用正确的AJAX格式在客户端处理响应，这是修改后的代码

      <div class="wrap-input100 validate-input" data-validate = "Enter username">
      <input class="input100" type="text" id="user" name="username" placeholder="Email">
      <span class="focus-input100" data-placeholder="&#xf207;"></span>
      </div>

      <div class="wrap-input100 validate-input" data-validate="Enter password">
      <input class="input100" type="password" id="pass" name="pass" placeholder="Password">
      <span class="focus-input100" data-placeholder="&#xf191;"></span>
      </div>

      <div class="container-login100-form-btn">
      <a class="login100-form-btn" id = "logBtn">
      Login
      </a>
      </div>
      <script src="https://code.jquery.com/jquery-2.2.4.min.js"></script>
      <script>
      $('#logBtn').click(function(event){ 
          user = document.getElementById("user").value;
          password = document.getElementById("pass").value;

          $.ajax({
              type:"POST",
              url:"login.php",
              async: false,
              data: {user:user,password:password},
              success: function(data){
              alert(data);
          if(data=="admin"){
                  window.location="https://..Main/index.html";
                }
        if(data=="user"){
                  window.location="https://....startemp.html";
                }
              }
              });
          });

      </script>

      <?php

      $servername = "localhost";
      $username = "root";
      $password = "root";
      $dbname = "test";

      $conn = new mysqli($servername, $username, $password, $dbname);

      $user = $_POST['user'];
      $pass = $_POST['password'];

      $sql = "SELECT * FROM users WHERE email='$user' AND clave='$pass'"; 

      $result = mysqli_query($conn, $sql);

      if (mysqli_num_rows($result) > 0) {
          $sql_1 = "SELECT * FROM users WHERE email='$user' AND clave='$pass' AND permisos='Administrador'";
           $result_1 = mysqli_query($conn, $sql_1);
          if (mysqli_num_rows($result_1) > 0){

              echo "admin";
              exit(0);
            }
             else{
          echo "user"; 
           exit(0);
          }

       } else {
          $msg = "username/password invalid";
          echo $msg;
       }

      mysqli_close($conn);
      ?>

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