jquery中的ajax如何返回结果而非回调方式即为同顺序执行

<br>function(url,params){ <br>var outdata; <br>$.ajax({ <br>type : "get", <br>async:false, <br>dataType:"json", <br>url : url, <br>data: params, <br>success : function(data){ <br>outdata = data; <br>}, <br>error:function(e){ <br>alert('ajax error'); <br>} <br>}); <br><br>return outdata; <br>} <br>