android 按两次返回键退出程序
单击返回键两次退出，单击一次返回键Toast提示,在其后2s内再单击一次返回键就退出。
方法一：
private static Boolean isExit =false;
private static Boolean hasTask =false;
Timer tExit =new Timer();
TimerTask task =new TimerTask() {
@Override
public void run() {
isExit =false;
hasTask =true;
}
};
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
// TODO Auto-generated method stub
if (keyCode == KeyEvent.KEYCODE_BACK) {
if(isExit ==false ) {
isExit =true;
Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();
if(!hasTask) {
tExit.schedule(task, 2000);
}
} else {
finish();
System.exit(0);
}
}
return false;
}
方法二：
private long mExitTime;
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK) {
if ((System.currentTimeMillis() - mExitTime) > 2000) {
Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();
mExitTime = System.currentTimeMillis();
} else {
finish();
}
return true;
}
return super.onKeyDown(keyCode, event);
}
}