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The Language Which does not need any prior knowledge of Programming and Easy to learn .Python is Object-oriented ,interpreted and Server side Scripting language . It is mainly used for lynux platform.
In Advance concept After learning Core Python We will use Python to create Desktop Application, Web Application, Sockets Programming , Multithread Programming. Since its An Open source Language its free of Cost
Ruby is server side, dynamic, reflective, object-oriented, general-purpose programming language. Ruby is "an interpreted scripting language for quick and easy object-oriented programming"
Ruby on Rails, or simply Rails, is a web application frameworkwritten in Ruby under the MIT License. Rails is a model–view–controller (MVC) framework, providing default structures for a database, a web service, and web pages.
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Step 1 Create an index.php file <form action="" class="form_registration" method="post"> <input type="text" name="first_name" value="" placeholder="first_name" /> <input type="text" name="last_name" value="" placeholder="last_name" /> <select name="gender"> <option value="male">Male</option> <option value="female">Female</option> </select> <input type="submit" name="submit" value="Insert" /> </form> <div class="result"> </div> <script type="text/javascript" src="jquery.js"></script> <script type="text/javascript"> $("document").ready(function() { $(".form_registration").submit(function() { var data = {"action": "test"}; data = $(this).serialize() + "&" + $.param(data); $.ajax({ type: "POST", dataType: "json", url: "response.php", data: data, success: function(data) { $(".result").html( "First name: " + data["first_name"] + "<br />last Name: " + data["last_name"] + "<br />Gender: " + data["gender"] + "<br />JSON: " + data["json"] ); } }); return false; }); }); </script> Step 2 create php server file to return data in json format <?php if (is_ajax()) { if (isset($_POST["action"]) && !empty($_POST["action"])) { $action = $_POST["action"]; switch($action) { case "test": test_function(); break; } } } function is_ajax() { return isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'; } function test_function() { $return = $_POST; $return["json"] = json_encode($return); echo json_encode($return); } ?>