如何從JSON數組中檢索數據？

Question

這裏是我的JSON數據

{ 
"list":[{"rank":"1","fb_id":"12345","score":"3000"}, 
     {"rank":"2","fb_id":"12345","score":"2000"},  
     {"rank":"3","fb_id":"123456","score":"1000"}, 
     {"rank":"5","fb_id":"123456","score":"2500"}], 
     "roundscores":[{"score1":"1000","pos1":"3","score2":"2500","pos2":"5"}] 
} 

從這個數據，我需要使用Java腳本獲得score1

我的JS

$.ajax({ 
type: "POST", 
url: "ajaxClassifica2.jsp", 
data: "selectedLevel="+selectedLevel+"&fb_id="+123456, 
success: function(data) { 
     alert(data); 
     var jsonlist = data.list; 
     var jsonround = data.roundscores; 
     alert(jsonlist); 
     alert(jsonround); 
     var strBuild = ""; 
     $.each(jsonobj, function(index, names) 
     { 
       strBuild ="<ul class=\"livelli\">" 
          +" <div class=\"social\"> " 
          +" <a href=\"#\"><img src=\"images/ico_fb.png\"></a>" 
          +" <a href=\"#\"><img src=\"images/ico_tw.png\"></a>" 
          +" </div>" 
          +" <li><strong>Round 1:</strong> " + names.roundscores.score1 + "</li>" 
          +" <li><strong>Posizione:</strong> " + names.roundscores.pos1 + " </li>" 
          +" </ul>" 
          +" <ul class=\"livelli\">" 
          +" <div class=\"social\"> " 
          +" <a href=\"#\"><img src=\"images/ico_fb.png\"></a>" 
          +"  <a href=\"#\"><img src=\"images/ico_tw.png\"></a>" 
          +" </div>" 
          +" <li><strong>Round 2:</strong> " + names.roundscores.score2 + "</li>" 
          +" <li><strong>Posizione:</strong> " + names.roundscores.pos2 + "</li>" 
          +" </ul>"; 

     }); 
       } 
    }); 

Answer 1

您可以使用JSON.parse然後用普通的JavaScript命名訪問：

var obj = JSON.parse('{"list":[...]}'); 
obj.roundscores[0].score1 // 1000 

另外，如果你有沒有在JSON.parse API舊的瀏覽器時，您可以使用填充工具如JSON3

Answer 2

var json = '{ "list":[{"rank":"1","fb_id":"12345","score":"3000"},{"rank":"2","fb_id":"12345","score":"2000"},{"rank":"3","fb_id":"123456","score":"1000"},{"rank":"5","fb_id":"123456","score":"2500"}], "roundscores":[{"score1":"1000","pos1":"3","score2":"2500","pos2":"5"}] }'; 
var obj = JSON.parse(json); 
var score1 = obj.roundscores[0].score1); 

Answer 3

免得假設你的JSON數據是變量「RAWDATA」。

var jData = JSON.parse(rawData); 
alert(jData.roundscores[0].score1); 

Answer 4

回顧這個例子可能是你

  var jsonText = '{"name":"Frodo","address":"Hobbiton, The Shire"}'; 
     var person = JSON.parse(jsonText); 

     alert(person.name);