Question : simple json example not working

Hi , it is an example from the book jquery 1.3 with php
but for some reason i can't get the callback function to work (this is what i think)


$.getJSON(
          'delete.php?ajax=true?id='+id,
           remove_row
          );


delete.php:

<?php
 $id = (int)$_REQUEST['id'];
  echo (!($id%2)) ?
  "{'id':$id,'success':1}":
  "{'id':$id,'success':0,'error':'Could not delete subscriber'}";

?>


  function remove_row(data)
   {  
    if (!data.success)  return alert(data.error);
    $('#delete_link_'+data.id)
    .closest('li')
    .remove();            
   }

if i call e.g. delete.php?id=6 alone it echoes back json data, but remove_row does not seem to be called

Answer : simple json example not working

The JSON being returned is incorrect. It currently is

{'id':$id,'success':1}

while it should be

{"id":$id,"success":1}

i.e. use double quotes for string instead of single quotes.