PHP-提交按钮，无需刷新即可获得价值

Question

Hello I want to get the value of this input and fetch it using ajax no database at all. thank you. how can i do it with ajax?

<form method="POST">
   <input type="text" name="input" id="card-code" value='<?php echo $code ?>' class="form-control">

   <input type="text" id="card-pin" value='<?php echo $code2 ?>' class="form-control" maxlength="3">
 </form>

there is my inputs and here is the button.

           <form action="top-up.php" method="POST"> 


                    </div>
                </div>

                <div class="col-md-6" style="margin-top: -160px">

                        <div class="caption">

                        <div class="jumbotron">

                        <textarea class="form-control text-center" id="scanned-QR" name="lblQrTxt" onchange = "change()"></textarea><br><br><br>

                           <input class="btn btn-primary btn-lg" type="submit" name="btnSubcode" value="PROCESS"></input>
                        </div>

                        </div>
                </div>
            </div>
             </form>

so the final output sould not refresh the page and the textarea value will be send to the textbox

Answer 1

The jQuery Form Plugin allows you to easily and unobtrusively upgrade HTML forms to use AJAX. The main methods, ajaxForm and ajaxSubmit, gather information from the form element to determine how to manage the submit process.

http://malsup.com/jquery/form/#getting-started

 $(document).ready(function() { // bind 'myForm' and provide a simple callback function $('#myForm').ajaxForm(function() { alert("Thank you for your comment!"); }); }); 

 <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script> <script src="http://malsup.github.com/jquery.form.js"></script> <form id="myForm" action="comment.php" method="post"> Name: <input type="text" name="name" /> Comment: <textarea name="comment"></textarea> <input type="submit" value="Submit Comment" /> </form> 

// prepare Options Object 
var options = { 
    target:     '#divToUpdate', 
    url:        'comment.php', 
    success:    function() { 
        alert('Thanks for your comment!'); 
    } 
}; 

// pass options to ajaxForm 
$('#myForm').ajaxForm(options);

Answer 2

Firstly, rewrite your html code as below:

<form id="form" action="top-up.php" method="POST"> 


                    </div>
                </div>

                <div class="col-md-6" style="margin-top: -160px">

                        <div class="caption">

                        <div class="jumbotron">

                        <textarea class="form-control text-center" id="scanned-QR" name="lblQrTxt"></textarea><br><br><br>

                           <input class="btn btn-primary btn-lg js-form-submit" type="submit"></input>
                        </div>

                        </div>
                </div>
            </div>
             </form>

Then, you can write JS something like this:

$(document).on('click','.js-form-submit', function (e) {
    e.preventDefault();
    var formData = $('#form').serialize();
    var url = $('#form').attr('action');
    $.ajax({
        type: "POST",
        cache: false,
        url: url // Your php url here
        data : formData,
        dataType: "json",
        success: function(response) {
             //var obj = jQuery.parseJSON(response); if the dataType is not specified as json uncomment this
             // do what ever you want with the server response
        },
        error: function() {
          alert('error handling here');
        }
    });
});