Question

我有一个闹钟，想要选择用户选择音乐。我有一些代码的活动，包括这个片段：

Button ring_button = (Button) findViewById(R.id.button2);
ring_button.setOnClickListener(new OnClickListener() {
    @Override
    public void onClick(View v) {
        // TODO Auto-generated method stub
        Intent intent_for_ring_button = new Intent();
        intent_for_ring_button.setAction(Intent.ACTION_GET_CONTENT);
        File file = new File(REPORTS_DIRECTORY);
        intent_for_ring_button.setDataAndType(Uri.fromFile(file),"audio/*");
        startActivityForResult(Intent.createChooser(intent_for_ring_button,"Open folder"), 0);
    }
});

还有一个带有闹钟的活动，扩展 BroadcastReceiver ，我开始下一个活动，带有一些音乐的对话窗口。这是它：

public class Dialog_window extends ActionBarActivity{

    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.dialog_window_bckg);
        getSupportActionBar().hide();
        this.getWindow().setSoftInputMode(WindowManager.LayoutParams.SOFT_INPUT_STATE_ALWAYS_HIDDEN);

        final MediaPlayer Alarm_mp3 = MediaPlayer.create(Dialog_window.this, R.raw.data_don_t_sing);
        Alarm_mp3.start();
        Alarm_mp3.setLooping(true);
        AlertDialog.Builder builder = new AlertDialog.Builder(Dialog_window.this);
        builder.setTitle("Важное сообщение!")
                .setMessage("Вставай!")
                .setIcon(R.drawable.uncle_sam)
                .setCancelable(false)
                .setNegativeButton("ОК, встаю, встаю.",
                        new DialogInterface.OnClickListener() {
                            public void onClick(DialogInterface dialog, int id) {
                                dialog.cancel();
                                Alarm_mp3.stop();
                                System.exit(0);
                            }
                        });
        AlertDialog alert = builder.create();
        alert.show();
    }

}

我怎么能意识到这一点？我必须从第一项活动中得到一些额外的东西或其他的东西吗？

Answer 1

使用var word = "aabb" let numberOfChars = word.characters.count // 4 let numberOfDistinctChars = Set(word.characters).count // 2 let occurrenciesOfA = word.characters.filter { $0 == "A" }.count // 0 let occurrenciesOfa = word.characters.filter { $0 == "a" }.count // 2 let occurrenciesOfACaseInsensitive = word.characters.filter { $0 == "A" || $0 == "a" }.count // 2 print(occurrenciesOfA) print(occurrenciesOfa) print(occurrenciesOfACaseInsensitive)意图调用startActivityForResult后，您应该覆盖同一类中的ACTION_GET_CONTENT，并获取所选文件的路径。获得路径后，您可以保存到共享首选项，并轻松从其他活动中检索（例如，在onActivityResult中）并使用Dialog_window进行游戏。

MediaPlayer

在另一项活动中检索后，只需简单地播放即可。

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    if(requestCode == 0){ // you've set to 0 at startActivityForResult 
        if(resultCode==RESULT_OK){
            String filePath = data.getData().getPath();
            // save filePath to shared preferences
       }
   }
 }

将文件路径保存到共享首选项：

String savedPath = ..... ; // retrieve from shared preferences  

//check your saved path, it could be a full path already..
//if it is, you don't need to concatenate anything to it
String fullPath = Environment.getExternalStorageDirectory()+savedPath;

mediaPlayer = new  MediaPlayer();
mediaPlayer.setDataSource(fullPath);
mediaPlayer.prepare();   
mediaPlayer.start()

从共享偏好设置恢复文件路径：

SharedPreferences.Editor editor = getSharedPreferences("mypref", MODE_PRIVATE).edit();
editor.putString("audioFilePath", filePath);
editor.commit();

从文件夹中获取音频

1 个答案: