如何獲取內容/字符串值，當我解析的網址？

Question

String url="http://graph.facebook.com/GHost"; 
URL objUrl = new URL(url); 
InputStream is= objUrl.openStream(); 
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); 
DocumentBuilder db = dbf.newDocumentBuilder(); 
Document doc = db.parse(new InputSource(is)); 
String str=doc.getTextContent();<<<<<<<<<----------- getting null value 
JSONObject object = new JSONObject(str);<<<<---------making exception error 
String id = object.getString("id"); 
if do open the url the content is like below 
{ 
    "id": "3965568744555", 
    "name": "GHost", 
    "picture": "http://profile.ak.fbcdn.net/hprofile-ak-snc4/373042_396504583708761_1625984_s.jpg", 
    "link": "http://www.facebook.com/GHost", 
    "likes": 70, 
    "cover": { 
    "cover_id": 493579484270, 
    "source": "http://a3.sphotos.ak.fbcdn.net/hphotos-ak-snc7/s720x720/396014_493579484001270_177795_n.jpg", 
    "offset_y": 0 
}, 

我該如何在String str =？編寫代碼？最後我想獲得id值。請任何幫助？

Answer 1

可以解析該文件......與使用concat功能..下面的代碼 parse json string

Answer 2

使用

String line; 
BufferedReader r = new BufferedReader(new InputStreamReader(inputStream)); 
StringBuilder total = new StringBuilder(); 
while ((line= r.readLine()) != null) { 
    total.append(line); 
} 

String str = total.toString(); 

，而不是

DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); 
DocumentBuilder db = dbf.newDocumentBuilder(); 
Document doc = db.parse(new InputSource(is)); 
String str=doc.getTextContent();