http://search.twitter.com/search.json?page=10&q=&geocode=37,-122,1mi&callback=myCallback
这是我的Twitter搜索网址。
如何获取响应对象?我知道如何解析一般的JSON对象,是否也是这样?
谢谢!
答案 0 :(得分:0)
尝试使用BufferedReader和StringBuilder(这就是我目前正在使用的应用程序)。然后我使用String来创建一个JSONObject。以下是您可能使用的一些模板代码:
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();
}