I want to change the view/Activity of my app after few seconds I mean i have created a home View for my app and i want to move to the next Activity after like 3 seconds, How should I achieve that. Thank You
问问题
169 次
4 回答
2
try this,
Handler mHandler = new Handler() {
public void handleMessage(android.os.Message msg) {
Intent i = new Intent(FirstActivity.this,SeconActivity.class);
startActivity(i);
};
};
mHandler.sendEmptyMessageDelayed(0, 3000);
于 2013-09-23T06:29:40.183 回答
0
view.postDelayed(Runnable r, int delay);
于 2013-09-23T06:29:27.127 回答
0
You can make slash Activity. Try this code....hop your problem will solve
public class SplashActivity extends Activity {
private final int SPLASH_DISPLAY_LENGHT = 2000;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.splash_screen);
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
/* Create an Intent that will start the Menu-Activity. */
Intent mainIntent = new Intent(SplashActivity.this, NightClubMain.class);
SplashActivity.this.startActivity(mainIntent);
SplashActivity.this.finish();
}
}, SPLASH_DISPLAY_LENGHT);
}
于 2013-09-23T06:30:25.467 回答
0
You can use Timer for that.
Timer myTimer;
startTimerTask();
public void startTimerTask() {
MyTimerTask myTask = new MyTimerTask();
myTimer = new Timer();
myTimer.schedule(myTask, 0, 3000);
}
@Override
public void onPause() {
super.onPause();
try {
myTimer.cancel();
} catch (Exception e) {
e.printStackTrace();
}
}
@Override
public void onStop() {
super.onStop();
try {
myTimer.cancel();
} catch (Exception e) {
e.printStackTrace();
}
}
class MyTimerTask extends TimerTask {
public void run() {
try {
getActivity().runOnUiThread(new Runnable() {
@Override
public void run() {
//
Do YOUR STUFF HERE
}
});
} catch (Exception e) {
e.printStackTrace();
}
}
}
于 2013-09-23T06:30:44.253 回答