How do I make my PHP AJAX code return false into my false jQuery block?

I have some jQuery that looks like this:

 $.ajax({
                type: "POST",
                url:开发者_JS百科 "/problems/vote.php",
                dataType: "json",
                data: dataString,
                success: function(json)
                {           
                    // ? :)



                }
                error : function() 
                {
                    alert("ajax error");
                }
            });

when I am writing my AJAX code, in case of some validation or database errors, I want to stop execution and return to the jQuery error block. What do I have to do in the PHP AJAX code in order to return things to the error block instead of the success block?

Thanks!!

---

you'd have to use the success function and capture what is returned for example:

 $.ajax({
            type: "POST",
            url: "/problems/vote.php",
            dataType: "json",
            data: dataString,
            success: function(json)
            {           
                if(json == "error"){
                   alert("error");
                }



            }
        });

you'll only use error: if there is a problem with the AJAX request