我最近在很多Android应用和游戏中都注意到了这种模式：当点击后退按钮“退出”应用程序时，Toast会出现类似“请再次单击BACK退出”的消息。
我想知道，因为我越来越频繁地看到它是一个内置的功能，你可以以某种方式访问活动？我查看了许多类的源代码，但我似乎无法找到任何相关内容。
当然，我可以考虑几种方法来轻松实现相同的功能（最容易的是在活动中保留一个布尔值，表明用户是否已经点击过一次......）但是我想知道这里是否有东西。
编辑：正如@LAS_VEGAS所说，我并不是指传统意义上的“退出”。 （即终止）我的意思是“回到应用程序启动活动启动之前打开的任何东西”，如果这是有意义的:)
在Java活动中：
boolean doubleBackToExitPressedOnce = false;
@Override
public void 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);
}
在Kotlin活动中：
private var doubleBackToExitPressedOnce = false
override fun onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed()
return
}
this.doubleBackToExitPressedOnce = true
Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show()
Handler().postDelayed(Runnable { doubleBackToExitPressedOnce = false }, 2000)
}
我认为这个处理程序有助于在2秒后重置变量。
它不是内置功能。我认为这甚至不是推荐的行为。 Android应用无意退出：
接受的答案是最好的，但如果您使用的是Android Design Support Library，那么您可以使用SnackBar获得更好的视图。
boolean doubleBackToExitPressedOnce = false;
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Snackbar.make(findViewById(R.id.photo_album_parent_view), "Please click BACK again to exit", Snackbar.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
}
@Override
public void onBackPressed() {
if (exitToast.getView().isShown()) {
exitToast.cancel();
finish();
} else {
exitToast.show();
}
}
这个工作正常，我已经测试过。我认为这更简单。
Zefnus使用System.currentTimeMillis（）的答案是最好的（+1）。我这样做的方式并不比这更好，但仍然发布它以增加上述想法。
如果按下后退按钮时看不到吐司，则会显示吐司，而如果可见（背面已在最后一个Toast.LENGTH_SHORT时间内按过一次），则退出。
exitToast = Toast.makeText(this, "Press again to exit", Toast.LENGTH_SHORT);
.
.
@Override
public void onBackPressed() {
if (exitToast.getView().getWindowToken() == null) //if toast is currently not visible
exitToast.show(); //then show toast saying 'press againt to exit'
else { //if toast is visible then
finish(); //or super.onBackPressed();
exitToast.cancel();
}
}
最近，我需要在我的应用程序中实现这个后退按钮功能。原始问题的答案很有用，但我还要考虑两点：
根据答案和评论，我创建了以下代码：
private static final long BACK_PRESS_DELAY = 1000;
private boolean mBackPressCancelled = false;
private long mBackPressTimestamp;
private Toast mBackPressToast;
@Override
public void onBackPressed() {
// Do nothing if the back button is disabled.
if (!mBackPressCancelled) {
// Pop fragment if the back stack is not empty.
if (getSupportFragmentManager().getBackStackEntryCount() > 0) {
super.onBackPressed();
} else {
if (mBackPressToast != null) {
mBackPressToast.cancel();
}
long currentTimestamp = System.currentTimeMillis();
if (currentTimestamp < mBackPressTimestamp + BACK_PRESS_DELAY) {
super.onBackPressed();
} else {
mBackPressTimestamp = currentTimestamp;
mBackPressToast = Toast.makeText(this, getString(R.string.warning_exit), Toast.LENGTH_SHORT);
mBackPressToast.show();
}
}
}
}
上面的代码假定使用了支持库。如果您使用片段但不使用支持库，则需要用getSupportFragmentManager()替换getFragmentManager()。
如果从未取消后退按钮，请删除第一个if。如果你不使用片段或片段后栈，删除第二个if
此外，重要的是要注意自Android 2.0以来支持onBackPressed方法。查看this page的详细说明。要使背压功能也适用于旧版本，请将以下方法添加到您的活动中：
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (android.os.Build.VERSION.SDK_INT < android.os.Build.VERSION_CODES.ECLAIR
&& keyCode == KeyEvent.KEYCODE_BACK
&& event.getRepeatCount() == 0) {
// Take care of calling this method on earlier versions of
// the platform where it doesn't exist.
onBackPressed();
}
return super.onKeyDown(keyCode, event);
}
在java中
private Boolean exit = false;
if (exit) {
onBackPressed();
}
@Override
public void onBackPressed() {
if (exit) {
finish(); // finish activity
} else {
Toast.makeText(this, "Press Back again to Exit.",
Toast.LENGTH_SHORT).show();
exit = true;
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
exit = false;
}
}, 3 * 1000);
}
}
在kotlin
private var exit = false
if (exit) {
onBackPressed()
}
override fun onBackPressed(){
if (exit){
finish() // finish activity
}else{
Toast.makeText(this, "Press Back again to Exit.",
Toast.LENGTH_SHORT).show()
exit = true
Handler().postDelayed({ exit = false }, 3 * 1000)
}
}
我知道这是一个非常古老的问题，但这是做你想做的最简单的方法。
@Override
public void onBackPressed() {
++k; //initialise k when you first start your activity.
if(k==1){
//do whatever you want to do on first click for example:
Toast.makeText(this, "Press back one more time to exit", Toast.LENGTH_LONG).show();
}else{
//do whatever you want to do on the click after the first for example:
finish();
}
}
我知道这不是最好的方法，但它工作正常！
当您将先前的堆栈活动存储在堆栈中时，这也会有所帮助。
我修改了Sudheesh的答案
boolean doubleBackToExitPressedOnce = false;
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
//super.onBackPressed();
Intent intent = new Intent(Intent.ACTION_MAIN);
intent.addCategory(Intent.CATEGORY_HOME);
intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);//***Change Here***
startActivity(intent);
finish();
System.exit(0);
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);
}
@Override public void onBackPressed() {
Log.d("CDA", "onBackPressed Called");
Intent intent = new Intent();
intent.setAction(Intent.ACTION_MAIN);
intent.addCategory(Intent.CATEGORY_HOME);
startActivity(intent);
}
为此，我实现了以下功能：
private long onRecentBackPressedTime;
@Override
public void onBackPressed() {
if (System.currentTimeMillis() - onRecentBackPressedTime > 2000) {
onRecentBackPressedTime = System.currentTimeMillis();
Toast.makeText(this, "Please press BACK again to exit", Toast.LENGTH_SHORT).show();
return;
}
super.onBackPressed();
}
Sudheesh B Nair在这个问题上有一个很好的（并且被接受的）答案，我认为应该有更好的选择，例如;
测量时间过去并检查自上次背压后TIME_INTERVAL毫秒（比如说2000）是否已经过去有什么问题。以下示例代码使用System.currentTimeMillis();来存储调用onBackPressed()的时间;
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();
}
回到接受的答案批评;使用flag来指示它是否在最后的TIME_INTERVAL（比如说2000）毫秒中被按下并设置 - 重置是通过Handler的postDelayed()方法是我想到的第一件事。但是当活动结束时，应该取消postDelayed()行动，删除Runnable。
为了删除Runnable，它不能被声明为匿名，并且与Handler一起被声明为成员。然后可以适当地调用removeCallbacks()的Handler方法。
以下示例是演示;
private boolean doubleBackToExitPressedOnce;
private Handler mHandler = new Handler();
private final Runnable mRunnable = new Runnable() {
@Override
public void run() {
doubleBackToExitPressedOnce = false;
}
};
@Override
protected void onDestroy()
{
super.onDestroy();
if (mHandler != null) { mHandler.removeCallbacks(mRunnable); }
}
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show();
mHandler.postDelayed(mRunnable, 2000);
}
感谢@NSouth的贡献;为了防止即使在应用程序关闭后出现toast消息，Toast也可以被声明为成员 - 比如mExitToast - 并且可以在mExitToast.cancel();调用之前通过super.onBackPressed();取消。
这是完整的工作代码。并且不要忘记删除回调，以免它在应用程序中导致内存泄漏。 :)
private boolean backPressedOnce = false;
private Handler statusUpdateHandler;
private Runnable statusUpdateRunnable;
public void onBackPressed() {
if (backPressedOnce) {
finish();
}
backPressedOnce = true;
final Toast toast = Toast.makeText(this, "Press again to exit", Toast.LENGTH_SHORT);
toast.show();
statusUpdateRunnable = new Runnable() {
@Override
public void run() {
backPressedOnce = false;
toast.cancel(); //Removes the toast after the exit.
}
};
statusUpdateHandler.postDelayed(statusUpdateRunnable, 2000);
}
@Override
protected void onDestroy() {
super.onDestroy();
if (statusUpdateHandler != null) {
statusUpdateHandler.removeCallbacks(statusUpdateRunnable);
}
}
在这里，我已经概括地编写了N个抽头计数的代码。该代码类似于Android设备手机中的Enable Developer选项编写。即使您可以在开发人员测试应用程序时使用此功能来启用功能。
private Handler tapHandler;
private Runnable tapRunnable;
private int mTapCount = 0;
private int milSecDealy = 2000;
onCreate(){
...
tapHandler = new Handler(Looper.getMainLooper());
}
在backpress或logout选项上调用askToExit（）。
private void askToExit() {
if (mTapCount >= 2) {
releaseTapValues();
/* ========= Exit = TRUE ========= */
}
mTapCount++;
validateTapCount();
}
/* Check with null to avoid create multiple instances of the runnable */
private void validateTapCount() {
if (tapRunnable == null) {
tapRunnable = new Runnable() {
@Override
public void run() {
releaseTapValues();
/* ========= Exit = FALSE ========= */
}
};
tapHandler.postDelayed(tapRunnable, milSecDealy);
}
}
private void releaseTapValues() {
/* Relase the value */
if (tapHandler != null) {
tapHandler.removeCallbacks(tapRunnable);
tapRunnable = null; /* release the object */
mTapCount = 0; /* release the value */
}
}
@Override
protected void onDestroy() {
super.onDestroy();
releaseTapValues();
}
当HomeActivity包含导航抽屉和双backPressed（）功能退出应用程序时。 （不要忘记初始化全局变量boolean doubleBackToExitPressedOnce = false;）2秒后新处理程序将doubleBackPressedOnce变量设置为false
@Override
public void onBackPressed() {
DrawerLayout drawer = findViewById(R.id.drawer_layout);
if (drawer.isDrawerOpen(GravityCompat.END)) {
drawer.closeDrawer(GravityCompat.END);
} else {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
moveTaskToBack(true);
return;
} else {
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);
}
}
}
boolean doubleBackToExitPressedOnce = false;
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Snackbar.make(findViewById(R.id.photo_album_parent_view), "Please click BACK again to exit", Snackbar.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
}
在Sudheesh B Nair的答案中有一些改进，我注意到它会等待处理程序，即使在立即按下两次，所以取消处理程序，如下所示。我也有tocled toast以防止它在app退出后显示。
boolean doubleBackToExitPressedOnce = false;
Handler myHandler;
Runnable myRunnable;
Toast myToast;
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
myHandler.removeCallbacks(myRunnable);
myToast.cancel();
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
myToast = Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT);
myToast.show();
myHandler = new Handler();
myRunnable = new Runnable() {
@Override
public void run() {
doubleBackToExitPressedOnce = false;
}
};
myHandler.postDelayed(myRunnable, 2000);
}
我觉得比Zefnus稍好一点的方法。只调用一次System.currentTimeMillis（）并省略return;：
long previousTime;
@Override
public void onBackPressed()
{
if (2000 + previousTime > (previousTime = System.currentTimeMillis()))
{
super.onBackPressed();
} else {
Toast.makeText(getBaseContext(), "Tap back button in order to exit", Toast.LENGTH_SHORT).show();
}
}
这与已接受和最多投票的回复相同，但是这个剪辑使用了Snackbar而不是Toast。
boolean doubleBackToExitPressedOnce = false;
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Snackbar.make(content, "Please click BACK again to exit", Snackbar.LENGTH_SHORT)
.setAction("Action", null).show();
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
}
在我的情况下，我依靠Snackbar#isShown()更好的UX。
private Snackbar exitSnackBar;
@Override
public void onBackPressed() {
if (isNavDrawerOpen()) {
closeNavDrawer();
} else if (getSupportFragmentManager().getBackStackEntryCount() == 0) {
if (exitSnackBar != null && exitSnackBar.isShown()) {
super.onBackPressed();
} else {
exitSnackBar = Snackbar.make(
binding.getRoot(),
R.string.navigation_exit,
2000
);
exitSnackBar.show();
}
} else {
super.onBackPressed();
}
}
对于具有导航抽屉的活动，请对OnBackPressed（）使用以下代码
boolean doubleBackToExitPressedOnce = false;
@Override
public void onBackPressed() {
DrawerLayout drawer = (DrawerLayout) findViewById(R.id.drawer_layout);
if (drawer.isDrawerOpen(GravityCompat.START)) {
drawer.closeDrawer(GravityCompat.START);
} else {
if (doubleBackToExitPressedOnce) {
if (getFragmentManager().getBackStackEntryCount() ==0) {
finishAffinity();
System.exit(0);
} else {
getFragmentManager().popBackStackImmediate();
}
return;
}
if (getFragmentManager().getBackStackEntryCount() ==0) {
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 {
getFragmentManager().popBackStackImmediate();
}
}
}
我用这个
import android.app.Activity;
import android.support.annotation.StringRes;
import android.widget.Toast;
public class ExitApp {
private static long lastClickTime;
public static void now(Activity ctx, @StringRes int message) {
now(ctx, ctx.getString(message), 2500);
}
public static void now(Activity ctx, @StringRes int message, long time) {
now(ctx, ctx.getString(message), time);
}
public static void now(Activity ctx, String message, long time) {
if (ctx != null && !message.isEmpty() && time != 0) {
if (lastClickTime + time > System.currentTimeMillis()) {
ctx.finish();
} else {
Toast.makeText(ctx, message, Toast.LENGTH_SHORT).show();
lastClickTime = System.currentTimeMillis();
}
}
}
}
用于事件onBackPressed
@Override
public void onBackPressed() {
ExitApp.now(this,"Press again for close");
}
或者ExitApp.now(this,R.string.double_back_pressed)
for change需要close，指定的毫秒数
ExitApp.now(this,R.string.double_back_pressed,5000)
我想我最后会分享我是如何做到的，我刚刚在我的活动中加入：
private boolean doubleBackToExitPressedOnce = false;
@Override
protected void onResume() {
super.onResume();
// .... other stuff in my onResume ....
this.doubleBackToExitPressedOnce = false;
}
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Toast.makeText(this, R.string.exit_press_back_twice_message, Toast.LENGTH_SHORT).show();
}
它完全符合我的要求。包括恢复活动时的状态重置。
这是另一种方法......使用CountDownTimer方法
private boolean exit = false;
@Override
public void onBackPressed() {
if (exit) {
finish();
} else {
Toast.makeText(this, "Press back again to exit",
Toast.LENGTH_SHORT).show();
exit = true;
new CountDownTimer(3000,1000) {
@Override
public void onTick(long l) {
}
@Override
public void onFinish() {
exit = false;
}
}.start();
}
}
流程图：
Java代码：
private long lastPressedTime;
private static final int PERIOD = 2000;
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (event.getKeyCode() == KeyEvent.KEYCODE_BACK) {
switch (event.getAction()) {
case KeyEvent.ACTION_DOWN:
if (event.getDownTime() - lastPressedTime < PERIOD) {
finish();
} else {
Toast.makeText(getApplicationContext(), "Press again to exit.",
Toast.LENGTH_SHORT).show();
lastPressedTime = event.getEventTime();
}
return true;
}
}
return false;
}
所有这些答案中都有最简单的方法。
只需在onBackPressed()方法中编写以下代码即可。
long back_pressed;
@Override
public void onBackPressed() {
if (back_pressed + 1000 > System.currentTimeMillis()){
super.onBackPressed();
}
else{
Toast.makeText(getBaseContext(),
"Press once again to exit!", Toast.LENGTH_SHORT)
.show();
}
back_pressed = System.currentTimeMillis();
}
您需要在活动中将back_pressed对象定义为long。
根据正确的答案和评论中的建议，我创建了一个完全正常的演示，并在使用后删除处理程序回调。
main activity.Java
package com.mehuljoisar.d_pressbacktwicetoexit;
import android.os.Bundle;
import android.os.Handler;
import android.app.Activity;
import android.widget.Toast;
public class MainActivity extends Activity {
private static final long delay = 2000L;
private boolean mRecentlyBackPressed = false;
private Handler mExitHandler = new Handler();
private Runnable mExitRunnable = new Runnable() {
@Override
public void run() {
mRecentlyBackPressed=false;
}
};
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
}
@Override
public void onBackPressed() {
//You may also add condition if (doubleBackToExitPressedOnce || fragmentManager.getBackStackEntryCount() != 0) // in case of Fragment-based add
if (mRecentlyBackPressed) {
mExitHandler.removeCallbacks(mExitRunnable);
mExitHandler = null;
super.onBackPressed();
}
else
{
mRecentlyBackPressed = true;
Toast.makeText(this, "press again to exit", Toast.LENGTH_SHORT).show();
mExitHandler.postDelayed(mExitRunnable, delay);
}
}
}
我希望它会有所帮助!!
我使用snackbar的解决方案：
Snackbar mSnackbar;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
final LinearLayout layout = findViewById(R.id.layout_main);
mSnackbar = Snackbar.make(layout, R.string.press_back_again, Snackbar.LENGTH_SHORT);
}
@Override
public void onBackPressed() {
if (mSnackbar.isShown()) {
super.onBackPressed();
} else {
mSnackbar.show();
}
}
简约时尚。
public void 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);
声明Variableprivate boolean doubleBackToExitPressedOnce = false;
将其粘贴到您的主要活动中，这将解决您的问题
退出应用程序时使用Runnable不是一个好主意，我最近想出了一种更简单的方法来记录和比较两个BACK按钮点击之间的时间。示例代码如下：
private static long back_pressed_time;
private static long PERIOD = 2000;
@Override
public void onBackPressed()
{
if (back_pressed_time + PERIOD > System.currentTimeMillis()) super.onBackPressed();
else Toast.makeText(getBaseContext(), "Press once again to exit!", Toast.LENGTH_SHORT).show();
back_pressed_time = System.currentTimeMillis();
}
这样就可以通过双重BACK按钮在一定的延迟时间内点击退出应用程序，该延迟时间为2000毫秒。