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Leaving android app with back button13个
我正在创建一个应用程序,我需要完成活动,当用户两次按下后退按钮.这是我尝试的代码
我正在创建一个应用程序,我需要完成活动,当用户两次按下后退按钮.这是我尝试的代码
@Override
public void onBackPressed() {
super.onBackPressed();
this.finish();
}
也试过这个
@Override
public boolean onKeyDown(int keyCode,KeyEvent event)
{
if ((keyCode == KeyEvent.KEYCODE_BACK))
{
finish();
}
return super.onKeyDown(keyCode,event);
}
这有助于我按下后退按钮完成活动.
请,我需要你的建议.提前致谢
解决方法
好的…这是一个更长但有效的方法来做到这一点……
1)在你的课堂上做一个全球可修复的…
private boolean backPressedToExitOnce = false; private Toast toast = null;
2)然后实现onBackPressed这样的活动……
@Override
public void onBackPressed() {
if (backPressedToExitOnce) {
super.onBackPressed();
} else {
this.backPressedToExitOnce = true;
showToast("Press again to exit");
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
backPressedToExitOnce = false;
}
},2000);
}
}
3)使用这个技巧有效地处理这种吐司…
/**
* Created to make sure that you toast doesn't show miltiple times,if user pressed back
* button more than once.
* @param message Message to show on toast.
*/
private void showToast(String message) {
if (this.toast == null) {
// Create toast if found null,it would he the case of first call only
this.toast = Toast.makeText(this,message,Toast.LENGTH_SHORT);
} else if (this.toast.getView() == null) {
// Toast not showing,so create new one
this.toast = Toast.makeText(this,Toast.LENGTH_SHORT);
} else {
// Updating toast message is showing
this.toast.setText(message);
}
// Showing toast finally
this.toast.show();
}
4)当你的活动关闭时,用这个技巧隐藏吐司……
/**
* Kill the toast if showing. Supposed to call from onPause() of activity.
* So that toast also get removed as activity goes to background,to improve
* better app experiance for user
*/
private void killToast() {
if (this.toast != null) {
this.toast.cancel();
}
}
5)实现你onPause()这样,当活动进入后台时杀死吐司
@Override
protected void onPause() {
killToast();
super.onPause();
}
希望这会有所帮助…… 原文链接：https://www.f2er.com/android/313819.html
