Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
------------------
$(document).ready(function(){
$("form#submit").submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: "POST",
url: "ajax.php",
data: "email="+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
------------------
And here is my json array in PHP:
------------------
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' => 1).
} else {
$json_data = array('result' => 0, 'error' => 'This is an error').
}
json_encode($json_data);
------------------
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb <falconwatc...@comcast.net> wrote:
> Return a JSON object.
>
> Construct a PHP array such as $json_data = array('result' => 0,
> 'error' => 'This is an error'). End your PHP script with json_encode
> ($json_data). Then you can reference del.result and del.error (I'm
> referring to your definition of the' success' function from your
> sample code)
>
> On Jan 4, 7:16 am, rob303 <r...@cube33.com> wrote:
>
> > Hi,
>
> > I'm new to ajax and jquery but I'm not new to PHP. The following
> > example seems to work okay to a point but I can't figure out how to
> > handle data validation errors generated in my PHP.
>
> > In my example I want to post some data to a script called ajax.php.
> > That script will check the data for validity and then return true or
> > false depending on the outcome of the checks. It will also set an
> > appropriate error message.
>
> > How can I handle the returned data and display an error message if
> > needed in ajax?
>
> > ----------------------
> > HTML / Ajax - test.php
> > ----------------------
> > <?php
>
> > require_once("includes.php");
>
> > ?>
> > <html>
> > <head>
> > <script src="js/jquery.js" type="text/javascript"></script>
> > <script type="text/javascript">
> > $(document).ready(function(){
> > $("form#submit").submit(function() {
> > var email = $('#email').attr('value');
> > $.ajax({
> > type: "POST",
> > url: "ajax.php",
> > data: "email="+ email,
> > success: function(del){
> > $('form#submit').hide();
> > $('div.success').fadeIn('medium');
> > // what if the data failed validation in PHP?
> > // we need to show 'div.error".
> > }
> > });
> > return false;
> > });
> > });
> > </script>
> > </head>
> > <body>
> > <form id="submit" method="post">
> > Email:
> > <input id="email" class="text" name="email" size="20"
> > type="text" />
> > <input type="submit" value="send mail" />
> > </form>
> > <div class="success" style="display:none;">
> > Email sent.
> > </div>
> > <div class="error">
> > <?php echo $userError; ?>
> > </div>
> > </body>
> > </html
>
> > ----------------------
> > PHP - ajax.php
> > ----------------------
> > <?php
>
> > require_once("includes.php");
>
> > $message = $_POST['email'];
>
> > if($message != '') {
> > mail('some-em...@email.com', 'Ajax test', $message);
> > return 1;} else {
>
> > $userError = "Please enter your email address";
> > return 0;
>
> > }
>
> > ?>
>
> > I've been searching the web for an answer but can't find one
> > anywhere! Any help would be greatly appreciated.
