2013-12-14 120 views
-1

我有这样一个奇怪的错误: 我试图解析来自url的json对象。这完美的作品,例如这是JSON数据:奇怪的json空指针

{"type":"result","rid":"djoezradio", 
"data":[{ 
"title":"Webradio", 
"song":"Test", 
"track":{ 
"artist":"Test", 
"title":"Test", 
"album":"", 
"Test":422, 
"id":423, 
"playlist":{ 
"id":14,"title":"reggae" 
}, 
"imageurl":"http:\/\/example.com\/static\/example\/covers\/nocover.png"}, 
"bitrate":"128 Kbps", 
"server":"Online","autodj":"Online","source":"Yes","offline":false","listeners":1, 
"maxlisteners":500,"reseller":0,"serverstate":true,"sourcestate":true, 
"sourceconn":true,"date":"Dec 14, 2013", 
"time":"02:13 PM","url":"http:\/\/example.com\/"}]} 

这是我的代码:

HttpClient client = new DefaultHttpClient(); 
    HttpGet request = new HttpGet(); 
    request.setURI(new URI("http://example.com")); 
    HttpResponse response; 
    try { 
     response = client.execute(request); 
     BufferedReader in = new BufferedReader(new InputStreamReader(response 
       .getEntity().getContent())); 
     String line = ""; 

     while ((line = in.readLine()) != null) { 

      JSONObject jObject = new JSONObject(line); 


       String temp = jObject.getString("imageurl"); 
       Log.e("rid",temp); 


     } 
    } catch (ClientProtocolException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } catch (IOException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } catch (JSONException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } 

当我做的getString(“播放列表”)为例,它只是运行良好,它会返回编号:14等 不工作的唯一的东西是对象imageurl ... 当我想解析这个,它只是返回null,而它只是在那里!

任何想法? 有什么理由吗?它因为它的一个.jpeg?恳求分享,我真的陷入困境。

+2

你确定正确张贴JSON的?它有不平衡的大括号。 – Henry

回答

1

试试这个。

HttpClient client = new DefaultHttpClient(); 
HttpGet request = new HttpGet(); 
request.setURI(new URI("http://example.com")); 
HttpResponse response; 
try { 
    response = client.execute(request); 
    BufferedReader in = new BufferedReader(new InputStreamReader(response 
      .getEntity().getContent())); 

StringBuilder builder = new StringBuilder(); 
String line; 
while((line=in.readLine())!=null) 
{ 

    builder.append(line); 
} 
String JSONdata = builder.toString(); 
Log.i("JsonData",JSONdata); 

JSONObject jObject = new JSONObject(JSONdata); 


String temp = jObject.getString("imageurl"); 
Log.e("rid",temp); 

} catch (ClientProtocolException e) { 
    // TODO Auto-generated catch block 
    e.printStackTrace(); 
} catch (IOException e) { 
    // TODO Auto-generated catch block 
    e.printStackTrace(); 
} catch (JSONException e) { 
    // TODO Auto-generated catch block 
    e.printStackTrace(); 
} 

编辑

JSONObject jObject = new JSONObject(JSONdata); 

JSONArray jdata = jObject.getJSONArray("data"); 

JSONObject job = jdata.getJSONObject(0); 
JSONObject jObjt = job.getJSONObject("track"); 
String temp = jObjt.getString("imageurl"); 
Log.e("rid",temp); 
+0

thnx,但仍然可以解析任何对象,但imageurl ....仍然返回null。 Log.e(“JsonData”,JSONdata);显示整个json数组,并在其中imageurl以及...仍然想知道为什么它返回null ..虽然它不是.. :( – iLuvCode

+0

@iLuvCode你需要从该JSON的数据是什么,也发布该响应也在log – Hariharan

+0

我需要的唯一数据是imageurl。当我做例如log.e(rid,jObject.getString(“playlist”))它只是起作用,但是当我想对imageurl做同样的事情时,imageurl返回null,而值只是像播放列表 – iLuvCode