jamie - 2:23 pm on Feb 1, 2012 (gmt 0)
put the php results in array format, then you can use json_encode() on it and return that.
you may find it easier to use $.ajax() instead of $.post() whilst learning as it forces you to write the parameters in longhand.
as long as you set the dataType as 'json', the data object which is returned is a json object. so if your original php array was array('result' => 'OK') then data.result contains OK.
the api docs explain it quite well: [api.jquery.com...]