Question

我想创建一个应用程序，让用户登录他的Youtube帐户，从他的帐户中选择一个播放列表，然后播放视频。

我正在尝试使用此文档连接到Youtube：

这是我在MainActivity中的相关代码（只是试图从YT API获得任何反应）：

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    // ...

    btnLogin = (Button)findViewById(R.id.btnlogin);
    btnLogin.setOnClickListener(new OnClickListener(){

        @Override
        public void onClick(View arg0) {

            Intent intent = AccountPicker.newChooseAccountIntent(null, null, new String[]{"com.google"},
                   false, null, null, null, null);
            startActivityForResult(intent, REQUEST_ACCOUNT_PICKER);
        }});
}

 @Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {        
    super.onActivityResult(requestCode, resultCode, data);
    switch (requestCode) {

        // ...

        case REQUEST_ACCOUNT_PICKER:

            if (resultCode == Activity.RESULT_OK && data != null
                    && data.getExtras() != null) {
                Log.i("VWL", "Callback");
                String accountName = data.getExtras().getString(
                        AccountManager.KEY_ACCOUNT_NAME);

                if (accountName != null) {                        

                    credential =
                            GoogleAccountCredential.usingOAuth2(this, Collections.singleton(YouTubeScopes.YOUTUBE_READONLY));

                    credential.setSelectedAccountName(accountName);

                    YouTube y = new YouTube.Builder(new NetHttpTransport(), new JacksonFactory(), credential).setApplicationName("VideoWhiteList").build();

                    try {
                        Log.i("VWL", "Playlist: " + y.playlists().list("snippet").setMine(true).execute().size());
                    } catch (IOException e) {
                        e.printStackTrace();
                    }                  

                    Log.i("VWL", "Account picked!");
                } else {
                    Log.i("VWL", "Something's fishy!");
                }
            }
            break;

    }
}

清单：

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
      package="x.y.z">
<uses-permission android:name="android.permission.INTERNET"/>
<uses-permission android:name="android.permission.GET_ACCOUNTS"/>
<uses-permission android:name="android.permission.USE_CREDENTIALS"/>
<uses-permission android:name="android.permission.MANAGE_ACCOUNTS"/>

<application
    android:allowBackup="true"
    android:icon="@mipmap/ic_launcher"
    android:label="@string/app_name"
    android:supportsRtl="true"
    android:theme="@style/AppTheme">

    <activity android:name="x.y.z.MainActivity"
              android:screenOrientation="landscape"
              android:configChanges="keyboardHidden|orientation|screenSize">
        <intent-filter>
            <action android:name="android.intent.action.MAIN"/>

            <category android:name="android.intent.category.LAUNCHER"/>
        </intent-filter>
    </activity>
</application>

出现AccountPicker，我可以选择我的帐户。之后，尝试通过y.playlists().行

访问YT API时出现以下错误

02-24 11:19:27.535 3379-3379/? E/ExceptionHandler: java.lang.RuntimeException: Failure delivering result ResultInfo{who=null, request=2, result=-1, data=Intent { (has extras) }} to activity {x.y.z/x.y.z.MainActivity}: java.lang.IllegalArgumentException: the name must not be empty: null

任何提示？

从Android访问YouTube API

0 个答案: