ajax中用josnp接收josn數據的實現方法

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);
 }
 },
 })
 });