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Showing posts with label ajax. Show all posts
Showing posts with label ajax. Show all posts

Call php function from ajax?

You can call PHP function like this using ajax, but not as directly call from AJAX url.

Here, Ajax Part  for explanation, set path url of db.php for call ajax and get result

 $.ajax({
      url: "http://localhost/db.php",
    data: frm_mail_data,
    cache: false,
    processData: false,
    contentType: false,
    type: 'POST',
        success: function (result) {
            alert(result);
            document.getElementById('div_result').innerHTML=result;
           }
  });

Step 1. Create index.php

Login form using php,Ajax and jquery with mysql

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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