如何從twitter搜索響應中解析JSON

Question

http://search.twitter.com/search.json?page=10&q=&geocode=37,-122,1mi&callback=myCallback

這是我的twitter搜索的網址。

如何獲取響應對象？我知道如何解析一般的JSON對象，它是一樣的嗎？

謝謝！

Answer 1

嘗試使用BufferedReader和StringBuilder（這就是我目前用於我的應用程序）。然後我使用String來創建一個JSONObject。下面是一些你可以使用模板代碼：

try{ 
    HttpClient httpclient = new DefaultHttpClient(); 
    HttpGet httpget = new HttpGet(//YOUR URL STRING HERE); 
    HttpResponse response; 
    response = httpclient.execute(httpget); 
    HttpEntity entity = response.getEntity(); 
    InputStream in = entity.getContent(); 

    BufferedReader reader = new BufferedReader(new InputStreamReader(in)); 
    StringBuilder sb = new StringBuilder(); 

    String input = null; 
    try { 
     while ((input = reader.readLine()) != null) { 
        sb.append(input + "\n"); 
     } 
    } catch (IOException e) { 
      e.printStackTrace(); 
    } finally { 
     try { 
      in.close(); 
     } catch (IOException e) { 
      e.printStackTrace(); 
     } 
    } 
    String enter = sb.toString(); 
     JSONObject obj = new JSONObject(enter); 

       //DO WHATEVER YOU WANT WITH THE JSONObject here 

     in.close(); 
} catch(MalformedURLException e){ 
    e.printStackTrace(); 
} catch (IOException e) { 
    e.printStackTrace(); 
} catch (JSONException e){ 
    e.printStackTrace(); 
}