Question

所以我在Android上制作了这个星球大战粉丝应用程序。

我有这个Yoda的形象和他的按钮。当你点击它时，现在他说一句话（与媒体播放器）。问题是，我希望他说不同的东西，所以我有4个不同的MP3文件，但是当用户点击按钮时，如何让它随机选择哪个播放？

这是我现在的代码：

package be.ehb.arnojansens.simpleFrag;

import android.app.Activity;
import android.media.MediaPlayer;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.Toast;

import be.ehb.arnojansens.fragmentexampleii.R;

public class SimpleFragmentActivity extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_simple_fragment);

        final Button advice = (Button) findViewById(R.id.YodaAdvice);
        final MediaPlayer mp = MediaPlayer.create(this, R.raw.yodamessage);

        advice.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {

                Toast.makeText(getApplicationContext(), "Hmm Busy I Am", Toast.LENGTH_LONG).show();
                mp.start();

            }
        });

    }

}

Answer 1

你可以有4条最终消息

final MediaPlayer mp = MediaPlayer.create(this, R.raw.yodamessage);
final MediaPlayer mp1 = MediaPlayer.create(this, R.raw.yodamessage1);//File names would be different I guess
final MediaPlayer mp2 = MediaPlayer.create(this, R.raw.yodamessage2);
final MediaPlayer mp3 = MediaPlayer.create(this, R.raw.yodamessage3);

你还需要随机

Random random=new Random();

然后在你的点击方法

int r = random.nextInt(4);
if(r==0){
mp.start();
Toast.makeText(getApplicationContext(), "Hmm Busy I Am", Toast.LENGTH_LONG).show();
} 
if(r==1){
mp1.start();
Toast.makeText(getApplicationContext(), "I'm hungry", Toast.LENGTH_LONG).show();
} 
if(r==2){
mp2.start();
Toast.makeText(getApplicationContext(), "My droid now", Toast.LENGTH_LONG).show();
} 
if(r==3){
mp3.start();
Toast.makeText(getApplicationContext(), "Not droid you are looking for", Toast.LENGTH_LONG).show();
}

随机音乐播放器onClick Android

1 个答案: