ajax juery debug

$.ajax({  
   type: "POST",  
   url: "post.php",  
   data: dataString,  
   success: function(data) {
       console.log('OK: ' + data);
   },
   error: function(jqXHR, textStatus, errorThrown) {
       console.log('Error: ' + errorThrown + ' ' + textStatus + ' ' + jqXHR);
   }
 });

借助于 console.log()函数