You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 <r...@cube33.com> wrote:
> 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.
