Thanks. I've installed the add-on. I'll give it a whirl.

On Apr 1, 8:00 pm, Hector Virgen <djvir...@gmail.com> wrote:
> You should try using Firebug. You can print directly to the Firebug console
> from within PHP, or you can just echo "some message" and look at the
> response tab of the firebug console.
> -Hector
>
> On Wed, Apr 1, 2009 at 11:56 AM, LinkGuru <i...@legalanalytics.co.uk> wrote:
>
> > Hi, I am looking for some help debugging jQuery being used with PHP in
> > the way described. I basically want a way of tracing where it gets to
> > in the PHP because the usual methods of writing out statements to the
> > page or embedding javascript alerts are not available to me in this
> > scenario. I will now explain.
>
> > I have some PHP (in a file called addToProj.php) which relies on the
> > last 3 lines of code being something similar to the following so that
> > jQuery can pick up returned data (including data for debugging
> > purposes)
>
> >                $json_array["status"]="point end";
> >                echo json_encode($json_array);
> > ?>
>
> > The jquery looks something like this, with the unimportant lines
> > removed.....
>
> >                $(".actionOptAdd").bind("click",
> >                        function()
> >                        {
> > .
> > .
> > .
> >                                $.getJSON('lib/addToProj.php',
> >                                {....}, function(data)
> >                        }, function(data)
> >                                {
> >                                   $.each(data, function(i,item)
> >                                   {
> >                                    if ( i == "status"
> > )$('#field).val(item);
> >                                   });
> >                                });
> >                        }
> >                );
> > //===========================
> > I am using json to return data to the jQuery (at the end of the day I
> > will need to return more than one data item) and then updating a field
> > on my html page. At the moment the return data is just for tracking
> > down where I have got to in the PHP. I should point out that I can't
> > insert statements like
>
> > echo "<script type='text/javascript'>alert('got here xx');</script>";
>
> > in my PHP (for some strange reason) to debug the PHP. Also, I can't
> > place lines like
>
> >                $json_array["status"]="message for debugging";
> >                echo json_encode($json_array);
> >                return();
>
> > at various points in the php, because it breaks the code. Also just
> > doing print '<p>something</p>'; can't be done in this case.
>
> > Has anyone got an idea of how I can solve the above. Your replies will
> > be greatly appreciated.
> > What is the easiest way of

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