Programming Techniques

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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);
     ?>

Jquery Tutorial 2: How can I know which radio button is selected via jQuery?

To get the value of the selected radioName item of a form with id myForm:

$('input[name=radioName]:checked', '#myForm').val()

Here's an example:

Index.html

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form id="myForm">
<input type="radio" name="radioName" value="1" /> 1 <br />
<input type="radio" name="radioName" value="2" /> 2 <br />
<input type="radio" name="radioName" value="3" /> 3 <br />
</form>

Javascript

$('#myForm input').on('change', function() {

alert($('input[name=radioName]:checked', '#myForm').val());
});

Here You can replace input class name and radio details

Programming Techniques

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Call php function from ajax?

Login form using php,Ajax and jquery with mysql

Jquery Tutorial 2: How can I know which radio button is selected via jQuery?

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