来自PHP的JQuery AJAX响应

I want to select data from database and show it in a div using PHP response. I am inserting data which is working fine for me but don't know what to do in PHP file for response and how to use it in AJAX done function. I am sharing my code here. Please help me if you can.

Script:

$(document).ready(function(){

$("#submit_btn").click(function(){

        var all_ok = 1;

        if ($("#course_name").val() === '') {
            $('#course_name_error').show();
            all_ok = 0;
        } else {
            $('#course_name_error').hide();
        }
        if (all_ok === 1) {
            var username = $("#username").val();
            var password = $("#password").val();

$.ajax({
  method: "POST",
  url: "login.php",
  data: { name: username, password: password }
})
  .done(function(data) {
    $("#success").text("Data added Successfully!").show();
});
}
});


});

HTML Part:

<form >
Name:<input type="name" name="name" id="username" />
<p id="username_error" style="color:red; display:none;">Enter a username.</p>
Password:<input type="password" name="password" id="password" />
<p id="username_error" style="color:red; display:none;">Enter a password.</p>
<button type="button" value="subbtn" id="submit_btn">Submit</button>
</form>

PHP:

<?php 

    $name = $_POST['name'];
    $pass = $_POST['password'];
    if((isset($name) && isset($pass))){

         include 'connection.php';
         $sql = "INSERT INTO `log_in` (id, name, password) VALUES (NULL, '$name', '$pass');";
         if (mysqli_query($conn, $sql)) {

         } 
         else {
           die( "Error: " . $sql . "<br>" . mysqli_error($conn));
         }
    }

Thanks in advance.