ajax中用josnp接收josn数据的实现方法

var queryUrl = "\"http://www.kuaidi100.com/query?type=" + $("#type").val() + "&postid=" + $("#postid").val() + "\"";
    $.ajax({
     url: 'http://query.yahooapis.com/v1/public/yql', //jsonp代理api
     dataType: "jsonp",
     data: {
      q: "select * from json where url=" + queryUrl,
      format: "json"
     },
     success: function(result) {
      var res = result.query.results.json;
      console.log(res);
      }
     },
    })
   });