oops.. forgot to change one line
CallParams.data = { imageid: image_id };
On Feb 27, 1:39 pm, MorningZ <morni...@gmail.com> wrote:
> instead of $.post, try this instead
>
> var CallParams = {};
> CallParams.type = "POST";
> CallParams.url = "your-page.php";
> CallParams.processData = true;
> CallParams.data = params;
> CallParams.dataType = "json";
> CallParams.success = function(data) {
> // "data" is the result};
>
> CallParams.error = function(x,y,z) {
> alert(x);};
>
> $.ajax(CallParams);
>
> that will show you what's going on
>
> On Feb 27, 12:41 pm, Dan <tribaldr...@gmail.com> wrote:
>
> > I'm using jquery 1.3.2 and trying to send some data using $.post to a
> > PHP script. For some reason the response is not being returned and
> > Firebug is reporting an error on line 3633 of jquery.
>
> > function check_out_image(image_id) {
> > $.post('ajax/checkout_image', {imageid: image_id} );
>
> > }
>
> > A very simply function that POST's an image ID to my PHP script. The
> > PHP script is getting run and updating a MySQL database so I know it's
> > calling the right function and working but the echo response never
> > makes it back.
>
> > For reference line 3633 of jquery 1.3.2 is the xhr.send(s.data) line
> > below:
>
> > try {
> > xhr.send(s.data);
> > } catch(e) {
> > jQuery.handleError(s, xhr, null, e);
> > }
>
> > For testing purposes I changed my PHP function to a simple echo "foo"
> > statement with nothing else in it and it still doesn't get returned.
> > Any ideas on why the response wouldn't get sent back?
>
> > Thanks,
> > Dan
