2012-03-16 77 views
1

我有一個JSON文件返回作爲JSON對象(它是一個數組的數組)...下面是返回的JSON對象如何顯示由JSON對象(的foreach道場)返回的數組

{ 
     "Info": { 
     "Contact": ".... ", 
     "title": "..." 
      }, 
     "details": [ 
      { 
       "ID": 1, 
       "Question": "User ID", 
       "Information": "", 

      }, { 
       "ID": 2, 
       "Question": "Name", 
       "Information": "", 

      }, { 
       "ID": 3, 
       "Question": "Age", 
       "Information": "", 

      } 
     ], 
    "list": [ 
      { 
       "No": 1, 
       "response": "" 
      }, { 
       "No": 2, 
       "response": "" 
      } 
     ] 
} 

現在我只想顯示細節......下面陣列

"Details": [ 
     { 
      "ID": 1, 
      "Question": "User ID", 
      "Information": "", 

     }, { 
      "ID": 2, 
      "Question": "Name", 
      "Information": "", 

     }, { 
      "ID": 3, 
      "Question": "Age", 
      "Information": "", 

     } 
    ], 

我如何做到這一點?請幫忙..

在此先感謝。

回答

1

1)解析JSON成一個JavaScript對象

var parsedJSON = JSON.parse(jsonData); 

2)訪問屬性你想

var details = parsedJSON.details; 

編輯:你正在分析你的JavaScript對象回JSON,爲什麼?

working jsfiddle

+0

我試圖這樣..詳情:功能(DObject){ 變種D進制= dojo.toJson(DObject); var D = DAry.details;的console.log(d);說:UNDEFINED ??/ – Newbie 2012-03-16 14:40:06

+0

@ user1257271你的JSON格式錯誤,你在每個細節對象的最後一個屬性上有一堆額外的逗號(,)。 – jbabey 2012-03-16 15:05:08

+0

@ user1257271請參閱我的答案中的編輯。你不應該做dojo.toJson – jbabey 2012-03-16 15:10:17

0
var output = ""; 
for(var i=0; i<json.details.length; i++) { 
    var detail = json.details[i]; 
    output += detail.ID +", "+ detail.Question +", "+ detail.Information +"\n"; 
} 
alert(output);