java - 关于无法使用Twitter开放api的问题？

最近需要做一个安卓app，主要实现获取他人@我的信息。
但是我遇到了一个问题。困扰了很久。
如果曾经做过，请帮助我一下。感激不尽！
我的QQ：514864610 多谢指教！多谢帮助！
我按照推特的要求，去申请了access_token：

随后我使用这个token，去实现对应的api，但是有些api请求到数据了：

但是有些api缺并没有请求到数据，而是返回了：

这是为什么呢？
根据官方的文档：
Bearer token used on endpoint which doesn’t support application-only auth
Requesting an endpoint which requires a user context (such as statuses/home_timeline) with a bearer token will produce:

HTTP/1.1 403 Forbidden
Content-Type: application/json; charset=utf-8
Content-Length: 91
...

{"errors":[{"message":"Your credentials do not allow access to this resource","code":220}]}

我不知道是否能放链接。放一个试试吧。这是我需要实现的请求
https://dev.twitter.com/rest/...
这是一个可以实现的请求：https://api.twitter.com/1.1/u...
附上一份我请求的代码
private void sendRequestWithHttpURLConnection() {

        
        new Thread(new Runnable() {
            @Override
            public void run() {

                HttpURLConnection connection = null;
                URL url = null;
                try {
                    url = new URL("https://api.twitter.com/1.1/statuses/mentions_timeline.json?count=2&since_id=14927799");
                    connection = (HttpURLConnection) url.openConnection();
                    connection.setRequestMethod("GET");
                    connection.setRequestProperty("Authorization","Bearer AAAAAAAAAAAAAAAAAAAAALrTxwAAAAAAYRuvnIXWe4pA8gPhRAuVsX2ND94%3DjcGVKktWDmLtCGwFYDoxayrqXwui6xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"这些是根据推特要求编码生成);
                    //详见https://dev.twitter.com/oauth/application-only

                    connection.setConnectTimeout(8000);
                    connection.setReadTimeout(8000);

// connection.connect();

                    InputStream in = connection.getInputStream();
                    BufferedReader reader = new BufferedReader(new InputStreamReader(in));
                    StringBuilder response = new StringBuilder();
                    String line;
                    while ((line = reader.readLine()) != null) {
                        response.append(line);
                    }
                    Message message = new Message();
                    message.what = SHOW_RESPONSE;
                    //将服务器返回的结果存放到Message中
                    message.obj = response.toString();
                    Log.d("Twitter","response.toString()="+response.toString());
                    handler.sendMessage(message);
                } catch (Exception e) {
                    e.printStackTrace();
                }finally {
                    if (connection!=null){
                        connection.disconnect();
                    }
                }
            }
        }).start();
    }