I wrote a piece of code that will give the user a prompt asking them to press back again if they would like to exit. I currently have my code working to an extent but I know it is written poorly and I assume there is a better way to do it. Any suggestions would be helpful!
Code:
public void onBackPressed(){
backpress = (backpress + 1);
Toast.makeText(getApplicationContext(), " Press Back again to Exit ", Toast.LENGTH_SHORT).show();
if (backpress>1) {
this.finish();
}
}
I just had this issue and solved it by adding the following method:
@Override
public boolean onOptionsItemSelected(MenuItem item) {
switch (item.getItemId()) {
case android.R.id.home:
// click on 'up' button in the action bar, handle it here
return true;
default:
return super.onOptionsItemSelected(item);
}
}
Use this, it may help.
@Override
public void onBackPressed() {
new AlertDialog.Builder(this)
.setTitle("Message")
.setMessage("Do you want to exit app?")
.setNegativeButton("NO", null)
.setPositiveButton("YES", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialogInterface, int i) {
UserLogin.super.onBackPressed();
}
}).create().show();
}
implementing onBackPressed() by System time, if pressed twice within 2 sec, then will exit
public class MainActivity extends AppCompatActivity {
private long backPressedTime; // for back button timing less than 2 sec
private Toast backToast; // to hold message of exit
@Override
public void onBackPressed() {
if (backPressedTime + 2000 > System.currentTimeMillis()) {
backToast.cancel(); // abruptly cancles the toast when pressed BACK Button *back2back*
super.onBackPressed();
} else {
backToast = Toast.makeText(getBaseContext(), "Press back again to exit",
Toast.LENGTH_SHORT);
backToast.show();
}
backPressedTime = System.currentTimeMillis();
}
}