Programming Techniques

Login form using php,Ajax and jquery with mysql, Redirect one to another page using ajax with return result.

Create a index.php and add this code to it


<html>
<body>
<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">
       <input class="input100" type="button" id="logBtn" name="logBtn" placeholder="Login" value="Login">
          </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/<file-name>.php";
               }
               if(data=="user"){
                 window.location="https://.....<file-name>.php";
               }
             }
             });
         });

     </script>
</body>
</html>

Create a Login.php

<?php

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

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

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

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

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

     if (mysqli_num_rows($result) > 0) {
         $sql_1 = "SELECT * FROM users WHERE email='$user' AND pass='$pass' AND type='admin'";
          $result_1 = mysqli_query($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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Login form using php,Ajax and jquery with mysql

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