的jQuery - 獲得可變

Question

Ajax響應和商店，我想存儲從http://www.youtube-mp3.org/a/itemInfo/?video_id=Vd85aPZ-QAE&ac=www&t=grp接收到的變量和訪問info.h數據..我想這

$.ajax({ 
      async: false, 
      type: "GET", 
      url: "http://www.youtube-mp3.org/a/itemInfo/?video_id=Vd85aPZ-QAE&ac=www&t=grp", 
      dataType: "json", 
      success : function(info) { 
         xml = info.h; 
       alert(xml); // WORKS 
        } 
     }); 

，但沒有成功..

infoRehashCallback=function(){4==h.readyState&&("$$$ERROR$$$"==h.responseText?pushItemError():500==h.status?(error_count+=1,4<error_count&&pushItemMaintenance()):(eval(h.responseText),checkInfo()))};infoRehash=function(){var a=new Date;h=createRequestObject();a="http://www.youtube-mp3.org/a/itemInfo/?video_id="+video_id+"&ac=www&t=grp&r="+a.getTime();h.onreadystatechange=infoRehashCallback;h.open("GET",a,!0);hs(h);h.send(null)}; 
startInfoRehash=function(a){video_id=a;infoRehash();interval=window.setInterval("infoRehash()",1E3*interval_diff)}; 

我從網站上的JavaScript的代碼

Answer 1

做出這樣的函數：

$.ajax({ 
      async: false, 
      type: "GET", 
      url: "http://www.youtube-mp3.org/a/itemInfo/?video_id=Vd85aPZ-QAE&ac=www&t=grp", 
      dataType: "json", 
      success : function(info) { 
         ShowAlert(info.h); 

        } 
     }); 

function ShowAlert(a) 
{ 
    alert(a); 
// do whatever want to with the result 
} 

Answer 2

我得到了以下回應這不是有效的JSON：

info = {"status" : "serving", "h" : "-"}; 

應該是：

{"status" : "serving", "h" : "-"}