Question

  $.ajax({  
            type: "POST",  
            url: "contacts.php",  
            data: dataString,  
            cache: false,  
            success: function(data, status, settings)  
            {  
               alert(The request URL and DATA);
            }  
            ,
            error: function(ajaxrequest, ajaxOptions, thrownError)  
            {  

            }  
        });

How can I alert the The request URL and DATA parameters inside the Success function?

Thank You

Answer 1

You can simply;

success: function(data, textStatus, jqXHR)
{
   alert(this.data + "," + this.url); 
}