You should really try the combination of Firebug and FirePHP...
http://www.firephp.org/
On Apr 2, 12:40 am, LinkGuru <i...@legalanalytics.co.uk> wrote:
> 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
