Question

我正在尝试用Android中的Xbox音乐API制作音乐播放器。到目前为止，我无法弄清楚如何准确检索访问令牌，如文档所述：

http://msdn.microsoft.com/en-us/library/dn546686.aspx

到目前为止，我有一个asynctask从我尝试使HttpPost工作（我是RESTful服务的新手）

这是我的代码：

@Override
protected String doInBackground(String... strings) {
    String postData = "client_id=" + Constants.CLIENT_ID
            + "&client_secret=" + Constants.CLIENT_SECRET
            + "&scope=" + Constants.SCOPE
            + "&grant_type=" + Constants.GRANT_TYPE;
    try {
        HttpClient client = new DefaultHttpClient();
        HttpPost request = new HttpPost(Constants.SERVICE + "/" + postData);
        HttpResponse response = client.execute(request);
        HttpEntity entity = response.getEntity();
        InputStream inputStream = entity.getContent();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"),8);
        String line = null;
        while((line=bufferedReader.readLine())!=null){
            Log.w(Constants.TAG, line);
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
    return null;
}

在Log.w上，我看到一个页面，显示＆＃34;服务器应用程序错误＆＃34;我不知道如何正确地做这个请求。

所有＆＃34; Constants.LIKE＆＃34;是文档说我需要的信息的字符串。

修改

这是我的常量（为我的应用安全而隐藏的秘密）

public static final String TAG="MUSIC_PLAYER_APP";
public static final String CLIENT_ID="musicplayer_internship_ldurazo";
public static final String CLIENT_SECRET="";
public static final String CALLBACK_URL="http://luisdurazoa.tumblr.com/";
public static final String SERVICE="https://datamarket.accesscontrol.windows.net/v2/OAuth2-13";
public static final String SCOPE="http://music.xboxlive.com";
public static final String TOKEN="TOKEN";
public static final String GRANT_TYPE="client_credentials";

Answer 1

如果有人遇到同样的问题，我修改了我的代码并解决了问题：

@Override
protected String doInBackground(String... strings) {
    try {
        StringBuilder stringBuilder = new StringBuilder();
        HttpClient client = new DefaultHttpClient();
        HttpPost request = new HttpPost(Constants.SERVICE);
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
        nameValuePairs.add(new BasicNameValuePair("client_id",Constants.CLIENT_ID));
        nameValuePairs.add(new BasicNameValuePair("client_secret",Constants.CLIENT_SECRET));
        nameValuePairs.add(new BasicNameValuePair("scope",Constants.SCOPE));
        nameValuePairs.add(new BasicNameValuePair("grant_type",Constants.GRANT_TYPE));
        request.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = client.execute(request);
        HttpEntity entity = response.getEntity();
        InputStream inputStream = entity.getContent();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"),8);
        String line = null;
        while((line=bufferedReader.readLine())!=null){
            stringBuilder.append(line);
            Log.w(Constants.TAG, line);
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
    return null;
}

缺少的是从＆＃34;行＆＃34;制作一个JSON对象。字符串，你很高兴。您可以使用我的代码在日志中查看您的访问令牌。

适用于Android的Xbox音乐API RESTful

1 个答案: