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Disable back button for previous activities, but exit with double pressed

Is there any way to disable in an activity the back button (to prevent show the previous activities), but also when you double press it, use it for exit??
I use this code that I found, but it only makes the second part. Close the app with double press..
What I try to do, is also prevent to go back and show the last visited activities. Thank you.
boolean doubleBackToExitPressedOnce = false;
#Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Toast.makeText(this, "Press again back for exit", Toast.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
#Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
}
UPDATE
I make the changes that you suggest, but now, it shows with the double back all the previous visited activities.
static final long THRESHOLD = 2000;
long backLastPressed;
boolean doubleBackToExitPressedOnce = false;
#Override
public void onBackPressed() {
if (System.currentTimeMillis() - backLastPressed < THRESHOLD) {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Toast.makeText(this, "Press again back for exit", Toast.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
#Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
backLastPressed = 0;
return;
}
backLastPressed = System.currentTimeMillis();
// Otherwise, ignore this BACK press
}

You could store an internal variable that marks the time of when the BACK button was pressed. Every time in onBackPressed check if the time now is within your double-tap threshold and if so, register it as a double-tap.
Try something like this:
static final long THRESHOLD = 2000;
long backLastPressed = 0;
#Override
public void onBackPressed() {
if (System.currentTimeMillis() - backLastPressed < THRESHOLD) {
// TODO: Register double-tapped BACK button, put your "exit" code here
backLastPressed = 0;
return;
}
backLastPressed = System.currentTimeMillis();
// Otherwise, ignore this BACK press
}
Don't call super.onBackPressed() if you're trying to override the BACK button behavior.

I still don't have the reputation to comment.
To prevent visiting the previous activities, you could, in addition to the code to do something when you double-tap back button, clear the tasks when you enter in this activity.
This guide shows how to achive this: https://developer.android.com/guide/components/activities/tasks-and-back-stack.html
TL;DR: use flags in the intent to clear the activity back stack.
In the intent:
Intent intent = new Intent(this, YourActivity.class);
intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK | Intent.FLAG_ACTIVITY_CLEAR_TASK);
startActivity(intent);
finish(); // call this to finish the previous activity

You can try my code :
private static long back_press;
#Override
public void onBackPressed() {
if (back_press + 2000 > System.currentTimeMillis()) {
super.onBackPressed();
} else {
Toast.makeText(getBaseContext(), "Please press again to exit!",
Toast.LENGTH_SHORT).show();
back_press = System.currentTimeMillis();
}
}
It worked for me, i hope it will help you.

Please, take a look at this sample app: https://github.com/IvanShafran/android-stackoverflow/tree/master/BackButtonHandle
Main point is:
private boolean mIsBackButtonPressedBefore = false
#Override
public void onBackPressed() {
if (mIsBackButtonPressedBefore) {
Toast.makeText(this, "Second back button press", Toast.LENGTH_SHORT).show();
super.onBackPressed();
} else {
mIsBackButtonPressedBefore = true;
Toast.makeText(this, "First back button press", Toast.LENGTH_SHORT).show();
}
}
It has described behaviour as you can see at the gif below:
If your code is similar to this but doesn't work you should look for a problem outside of this scope.

how can i add multiple click on back pressed button

How can I add multiple backpressed button like when i press back button that show me PRESS DOUBLE CLICK TO EXIT after toast message when i click double on press button then the app exit... i did this method but its just twice to exit the app i want to implement 3 times
if (doubleBackToExitPressedOnce){
super.onBackPressed();
doubleBackToExitPressedOnce = false;
}
else {
doubleBackToExitPressedOnce = true;
Toast.makeText(this, "Double Press to exit", Toast.LENGTH_SHORT).show();
}

I would suggest that way:
int counter = 0;
....
public void onBackPressed() {
counter++;
if(counter > 2){
System.exit(0);
}else{
Toast.makeText(this, "TRIPLE CLICK TO EXIT!", Toast.LENGTH_SHORT).show();
}
final long DELAY_TIME = 3000L;
new Thread(new Runnable() {
public void run(){
try {
Thread.sleep(DELAY_TIME);
counter = 0;
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}).start();
}

Well that seems super hacky from a UI perspective, but if you really want to do that an easy way might be something like:
Integer backToExitPressedCounter = 0
...
if(backToExitPressedCounter==3){
super.onBackPressed();
backToExitPressedCounter = 0;
} else {
backToExitPressedCounter++;
Toast.makeText(this, "Triple Press to exit", Toast.LENGTH_SHORT).show();
}

Activity closed before pressing back button twice

I want to close application after pressing back button twice. I have diffrent fragments to load on the activity. I want to close the application only when i am on the main(Home) fragment and user pressed back button twice. But it is not working properly. Here is my code to close the activity :-
#Override
public void onBackPressed() {
if (getFragmentManager().getBackStackEntryCount() == 0) {
//super.onBackPressed();
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
#Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
} else {
backFlag = 1;
getFragmentManager().popBackStack();
}
It is giving me message of "Press back button twice" but fragment is no transfer to main fragment. Please help me to solve this
Edited question
if (getFragmentManager().getBackStackEntryCount() == 0) {
super.onBackPressed();
} else {
getFragmentManager().popBackStack();
}

You can achieve this by trying this alternative method by keeping two second interval to press back again:
private static final int TIME_INTERVAL = 2000; // # milliseconds, desired time passed between two back presses.
private long mBackPressed;
#Override
public void onBackPressed()
{
if (mBackPressed + TIME_INTERVAL > System.currentTimeMillis())
{
super.onBackPressed();
return;
}
else { Toast.makeText(getBaseContext(), "Tap back button in order to exit", Toast.LENGTH_SHORT).show(); }
mBackPressed = System.currentTimeMillis();
}

Using onbackpressed on homepage to exit app, setting homepage variable

I'm developing an Android Webview app. I would like to use onbackpressed to go to the previous page but also to exit the app.
How it should work:
Pressing back-button on homepage: I want to show a message "Press once again to exit."
Pressing back-button on other webpages: it should immediately open the previous page.
I tried to achieve this by using the following code. The only problem is that I don't know how to set my homepage as a variable. Right now I'm getting the error message "Page cannot be resolved as a variable". I understand why I get that message, I just don't know how to solve it in this particular case.
Let's say my homepage url is "http://example.com", how do I set this homepage as a variable?
private Boolean exit = false;
#Override
public void onBackPressed() {
if(page != "homepage"){
super.onBackPressed(); // Calls the Overriden Method
}
else
{
if (exit)
this.finish();
else {
Toast.makeText(this, "Press once again to exit.",
Toast.LENGTH_SHORT).show();
exit = true;
new Handler().postDelayed(new Runnable() {
#Override
public void run() {
exit = false;
}
}, 3 * 1000);
}
}
}

You can check isback page for webview then you can write down the code for exist from app. You can use following code for check
if(webView.canGoBack())
{
// your code
}

maybe this will help:
int clickedTwice = 0;
public void onBackPressed() {
if(page != "homepage"){
super.onBackPressed(); // Calls the Overriden Method
}
else
{
clickedTwice++;
if(clickedTwice == 2)
{
super.onBackPressed();
}
else
{
Toast.makeText(this,"press again to exit",Toast.LENGTH_SHORT).show();
}
new Handler().postDelayed(new Runnable()
{
#Override
public void run()
{
clickedTwice--;
}
}, 2000); // will wait for 2 seconds for user's another click!
}
}
}

Android: Proper Way to use onBackPressed() with Toast

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 would implement a dialog asking the user if they wanted to exit and then call super.onBackPressed() if they did.
#Override
public void onBackPressed() {
new AlertDialog.Builder(this)
.setTitle("Really Exit?")
.setMessage("Are you sure you want to exit?")
.setNegativeButton(android.R.string.no, null)
.setPositiveButton(android.R.string.yes, new OnClickListener() {
public void onClick(DialogInterface arg0, int arg1) {
WelcomeActivity.super.onBackPressed();
}
}).create().show();
}
In the above example, you'll need to replace WelcomeActivity with the name of your activity.

You don't need a counter for back presses.
Just store a reference to the toast that is shown:
private Toast backtoast;
Then,
public void onBackPressed() {
if(USER_IS_GOING_TO_EXIT) {
if(backtoast!=null&&backtoast.getView().getWindowToken()!=null) {
finish();
} else {
backtoast = Toast.makeText(this, "Press back to exit", Toast.LENGTH_SHORT);
backtoast.show();
}
} else {
//other stuff...
super.onBackPressed();
}
}
This will call finish() if you press back while the toast is still visible, and only if the back press would result in exiting the application.

I use this much simpler approach...
public class XYZ extends Activity {
private long backPressedTime = 0; // used by onBackPressed()
#Override
public void onBackPressed() { // to prevent irritating accidental logouts
long t = System.currentTimeMillis();
if (t - backPressedTime > 2000) { // 2 secs
backPressedTime = t;
Toast.makeText(this, "Press back again to logout",
Toast.LENGTH_SHORT).show();
} else { // this guy is serious
// clean up
super.onBackPressed(); // bye
}
}
}

Both your way and #Steve's way are acceptable ways to prevent accidental exits.
If choosing to continue with your implementation, you will need to make sure to have backpress initialized to 0, and probably implement a Timer of some sort to reset it back to 0 on keypress, after a cooldown period. (~5 seconds seems right)

You may also need to reset counter in onPause to prevent cases when user presses home or navigates away by some other means after first back press. Otherwise, I don't see an issue.

If you want to exit your application from direct Second Activity without going to First Activity then try this code..`
In Second Activity put this code..
#Override
public void onBackPressed() {
new AlertDialog.Builder(this)
.setTitle("Really Exit?")
.setMessage("Are you sure you want to exit?")
.setNegativeButton(android.R.string.no, null)
.setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
public void onClick(DialogInterface arg0, int arg1) {
setResult(RESULT_OK, new Intent().putExtra("EXIT", true));
finish();
}
}).create().show();
}
And Your First Activity Put this code.....
public class FirstActivity extends AppCompatActivity {
Button next;
private final static int EXIT_CODE = 100;
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
next = (Button) findViewById(R.id.next);
next.setOnClickListener(new View.OnClickListener() {
#Override
public void onClick(View view) {
startActivityForResult(new Intent(FirstActivity.this, SecondActivity.class), EXIT_CODE);
}
});
}
#Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == EXIT_CODE) {
if (resultCode == RESULT_OK) {
if (data.getBooleanExtra("EXIT", true)) {
finish();
}
}
}
}
}

This is the best way, because if user not back more than two seconds then reset backpressed value.
declare one global variable.
private boolean backPressToExit = false;
Override onBackPressed Method.
#Override
public void onBackPressed() {
if (backPressToExit) {
super.onBackPressed();
return;
}
this.backPressToExit = true;
Snackbar.make(findViewById(R.id.yourview), getString(R.string.exit_msg), Snackbar.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
#Override
public void run() {
backPressToExit = false;
}
}, 2000);
}

additionally, you need to dissmis dialog before calling activity.super.onBackPressed(), otherwise you'll get "Activity has leaked.." error.
Example in my case with sweetalerdialog library:
#Override
public void onBackPressed() {
//super.onBackPressed();
SweetAlertDialog progressDialog = new SweetAlertDialog(this, SweetAlertDialog.WARNING_TYPE);
progressDialog.setCancelable(false);
progressDialog.setTitleText("Are you sure you want to exit?");
progressDialog.setCancelText("No");
progressDialog.setConfirmText("Yes");
progressDialog.setCanceledOnTouchOutside(true);
progressDialog.setConfirmClickListener(new SweetAlertDialog.OnSweetClickListener() {
#Override
public void onClick(SweetAlertDialog sweetAlertDialog) {
sweetAlertDialog.dismiss();
MainActivity.super.onBackPressed();
}
});
progressDialog.show();
}

use to .onBackPressed() to back Activity specify
#Override
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);
}
}

You can also use onBackPressed by following ways using customized Toast:
enter image description here
customized_toast.xml
<?xml version="1.0" encoding="utf-8"?>
<TextView
xmlns:android="http://schemas.android.com/apk/res/android"
android:id="#+id/txtMessage"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:drawableStart="#drawable/ic_white_exit_small"
android:drawableLeft="#drawable/ic_white_exit_small"
android:drawablePadding="8dp"
android:paddingTop="8dp"
android:paddingBottom="8dp"
android:paddingLeft="16dp"
android:paddingRight="16dp"
android:gravity="center"
android:textColor="#android:color/white"
android:textSize="16sp"
android:text="Press BACK again to exit.."
android:background="#drawable/curve_edittext"/>
MainActivity.java
#Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
android.os.Process.killProcess(Process.myPid());
System.exit(1);
return;
}
this.doubleBackToExitPressedOnce = true;
Toast toast = new Toast(Dashboard.this);
View view = getLayoutInflater().inflate(R.layout.toast_view,null);
toast.setView(view);
toast.setDuration(Toast.LENGTH_SHORT);
int margin = getResources().getDimensionPixelSize(R.dimen.toast_vertical_margin);
toast.setGravity(Gravity.BOTTOM | Gravity.CENTER_VERTICAL, 0, margin);
toast.show();
new Handler().postDelayed(new Runnable() {
#Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
}

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();
}
}