在我的程序中,我只有1個按鈕。首先按程序會輸出一個隨機字符串。如果用戶再次按下它停止,我的程序將隨機(延遲)相同的老虎機。我該怎麼做?如何隨機字符串看起來像老虎機
我的代碼
package com.Randomsentence;
import java.util.Random;
import android.app.Activity;
import android.content.res.Resources;
import android.os.Bundle;
import android.os.Handler;
import android.os.Message;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
public class Randomsentence extends Activity {
boolean showRandom = false;
TextView txt;
int time = 30;
int random;
public String[] myString;
Button bt1;
boolean check = false;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
txt = (TextView) findViewById(R.id.txt);
bt1 = (Button) findViewById(R.id.bt1);
Medaiplayer mp = new Medaiplayer();
Mediaplayer mp2 = new Mediaplayer();
mp = MediaPlayer.create(getApplicationContext(), R.raw.AudioFile1);
mp2 = MediaPlayer.create(getApplicationContext(), R.raw.AudioFile2);
bt1.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
// TODO Auto-generated method stub
showRandom = !showRandom;
t = new Thread() {
public void run() {
try {
while (showRandom) {
mp.start();
mp2.reset();
mp2.prepare();
sleep(1000);
handler.sendMessage(handler.obtainMessage());
}
mp.reset();
mp.prepare();
mp2.start();
} catch (Exception ex) {
ex.printStackTrace();
}
}
};
t.start();
}
});
}
// our handler
Handler handler = new Handler() {
public void handleMessage(Message msg) {//display each item in a single line
{
Random rgenerator = new Random();
Resources res = getResources();
myString = res.getStringArray(R.array.myArray);
String q = myString[rgenerator.nextInt(myString.length)];
txt.setText(q);
}
}
};
}
'慢隨機(延遲)相同lottery'沒有意義。 – Atreys
這意味着當我再次按下按鈕時,字符串不會立即停止。但它會減緩停止類似的電影 – user814154
你是什麼意思的「彩票」?據我從代碼可以看出,這與彩票無關。 –