Question

我之前曾问过这个问题，但遗憾的是我无法找到关于这个问题的更好的解决方案。现在我的应用程序包含第一个屏幕作为启动画面，然后是列表视图（主要活动），然后单击列表视图的每一行打开每个活动我的要求是，如果我们从我的任何内部活动（当我们点击列表视图行时开放的活动）按下一次后退按钮，它必须导航到我的主列表视图然后如果我们再次从列表视图中按下该应用程序必须如果我按下我的列表视图中的后退按钮两次它将正确退出应用程序。但我的主要问题是，如果我从我的任何内部活动按两次后退按钮我的应用程序没有关闭我需要按三次，而不是从我的任何内部活动中关闭应用程序。任何人都可以帮助我吗？

我退出应用程序的代码就是这个我在主listview类中添加了这段代码..

private static final int TIME_INTERVAL = 3000; // # 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();
}
}

我的manifest.xml

<application
    android:allowBackup="true"
    android:label="@string/app_name"
    android:theme="@style/AppTheme"> 

    <meta-data android:name="com.google.android.gms.version"
       android:value="@integer/google_play_services_version"/>


    <activity
        android:name="learnersseries.mathematics.complexnumbers.Firstintro"
         android:screenOrientation="portrait" 
         android:launchMode="singleTop"            
        android:label="@string/app_name" >


        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />

        </intent-filter>
    </activity>


    <activity android:name="Myintegralpage"
        android:screenOrientation="portrait"
                   >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="myimagine"
        android:screenOrientation="portrait" 

       >
        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Myintroductionpage"
        android:screenOrientation="portrait" 

      >
        <intent-filter></intent-filter>
    </activity>
    <activity android:name="MainActivity"
        android:noHistory="false"
        android:screenOrientation="portrait" 
       >


        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Complexnumbers"
        android:screenOrientation="portrait"
         >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Equality"
        android:screenOrientation="portrait"
         >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Additionofcomplex"
        android:screenOrientation="portrait"
          >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Subtraction"
        android:screenOrientation="portrait" 
        >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="multiplication"
        android:screenOrientation="portrait" >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Division"
        android:screenOrientation="portrait" >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Conjugate"
        android:screenOrientation="portrait" >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Modulus"
        android:screenOrientation="portrait" >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Reciprocal"
        android:screenOrientation="portrait" >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Square"
        android:screenOrientation="portrait">


    </activity>
    <activity android:name="Representation"
        android:screenOrientation="portrait" >

        <intent-filter></intent-filter>
    </activity>
    <activity android:name="Argument"
        android:screenOrientation="portrait" >

Answer 1

请尝试这种方式，希望这有助于您解决问题。

在你的Activity中取一个标志doubleBackToExitPressedOnce默认（false），当第一次按下后退按钮时，标志值变为（ture），如果再次没有按下后退按钮，则在你应用程序的2秒内再次按下后退按钮2秒设置标志值（false）。

private boolean backPressedToExitOnce;

@Override
public void onBackPressed() {
    if (doubleBackToExitPressedOnce) {
        super.onBackPressed();
        return;
    }

    this.doubleBackToExitPressedOnce = true;
    Toast.makeText(this, "Press again to exit", Toast.LENGTH_SHORT).show();

    new Handler().postDelayed(new Runnable() {

        @Override
        public void run() {
            doubleBackToExitPressedOnce=false;                       
        }
    }, 2000);
}

Answer 2

您好我知道我的答案为时已晚，但请仔细阅读以下代码段以获得明确的解决方案：

@Override
public void onBackPressed() {
    if(canExit)
        super.onBackPressed();
    else{
        canExit = true;
        Toast.makeText(getApplicationContext(), "Press again", Toast.LENGTH_SHORT).show();
    }
    mHandler.sendEmptyMessageDelayed(1, 2000/*time interval to next press in milli second*/);// if not pressed within 2seconds then will be setted(canExit) as false
}


public Handler mHandler = new Handler(){

        public void handleMessage(android.os.Message msg) {

            switch (msg.what) {
            case 1:
                canExit = false;
                break;
            default:
                break;
            }
        }
    };

Answer 3

这是执行此操作的完整工作和简单代码。并且不要忘记删除onDestroy方法中的回调，以便它不会在应用程序中导致内存泄漏。 :)

private boolean backPressedOnce = false;
private Handler statusUpdateHandler = new Handler();
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);
    }
}

Answer 4

这是一个有保障的工作解决方案，可以按2次退出应用

int doubleBackToExitPressed = 1;
@RequiresApi(api = Build.VERSION_CODES.JELLY_BEAN)
@Override
public void onBackPressed() {
    if (doubleBackToExitPressed == 2) {
        finishAffinity();
        System.exit(0);
    }
    else {
        doubleBackToExitPressed++;
        Toast.makeText(this, "Please press Back again to exit", Toast.LENGTH_SHORT).show();
    }

    new Handler().postDelayed(new Runnable() {
        @Override
        public void run() {
            doubleBackToExitPressed=1;`enter code here`
        }
    }, 2000);
}

Answer 5

尝试将“父活动”添加到清单中的新活动例如：

<activity android:name="myimagine"
          android:screenOrientation="portrait"
          android:parentActivityName="com.example.previous_activity" />

Answer 6

添加到Haresh Chhelana ...只需创建一个基本活动，然后从中扩展所有其他活动。由Haresh Chhelana编写的代码插入到BaseActivity中。我认为它必须解决你的问题

Answer 7

当您覆盖onBackPressed时，在第二次点击后您需要：

完成应用程序流程：

ActivityManager am =（ActivityManager）getSystemService（Activity.ACTIVITY_SERVICE）; am.killBackgroundProcesses（的packageName）;
完成您的第一项活动。以下是示例：https://stackoverflow.com/a/14002030/3864698

<强> UPD

第一种情况：

@Override
    public void onBackPressed() {
        if (doubleBackToExitPressedOnce) {
            ActivityManager manager =  (ActivityManager) this.getSystemService(ACTIVITY_SERVICE);
            List<ActivityManager.RunningAppProcessInfo> activityes = ((ActivityManager) manager).getRunningAppProcesses();
            for (int i = 0; i < activityes.size(); i++){
                if (activityes.get(i).processName.equals(getApplicationInfo().packageName)) {
                    android.os.Process.killProcess(activityes.get(i).pid);
                }
            }
            return;
        }
        doubleBackToExitPressedOnce = true;


        new Handler().postDelayed(new Runnable() {

            @Override
            public void run() {
                doubleBackToExitPressedOnce = false;
                myOwnBackPress();
                }
        }, 1000);
}

private void myOwnBackPress() {
    if(!isFinishing()) {
        super.onBackPressed();
    }
}

实际上谷歌不建议杀死进程。

第二种情况：

//for all activity besides HomeActivity.
@Override
    public void onBackPressed() {
        if (doubleBackToExitPressedOnce) {
            Intent intent = new Intent(getApplicationContext(), HomeActivity.class);
            intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
            intent.putExtra("EXIT", true);
            startActivity(intent);
            return;
        }
        doubleBackToExitPressedOnce = true;


        new Handler().postDelayed(new Runnable() {

            @Override
            public void run() {
                doubleBackToExitPressedOnce = false;
                myOwnBackPress();
            }
        }, 1000);
}

private void myOwnBackPress() {
    if(!isFinishing()) {
        super.onBackPressed();
    }
}

在您的HomeActivity中，不要覆盖onBackPressed并在onCreate中添加下一个：

 @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
       // initialize a lot of variables
        if (getIntent().getBooleanExtra("EXIT", false)) {
            finish();
        }
        // some code..
}

Answer 8

试试这种方式，希望能帮到你！

    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) {
                moveTaskToBack(true);
                android.os.Process.killProcess(android.os.Process.myPid());
                System.exit(1);
            } else {
                Toast.makeText(getApplicationContext(),
                        "Press again to exit.", Toast.LENGTH_SHORT).show();
                lastPressedTime = event.getEventTime();
            }
            return true;
        }
    }
    return false;
}

Answer 9

在不得不多次实现捕获双重背压动作的行为要求之后，构建了一个库就可以做到这一点。 DoubleBackPress Android Library使用内置模板轻松处理此类情况。

因此，对于像退出按钮两次退出一样，只需执行：

// set the Action to occur on DoubleBackPress
DoubleBackPressAction doubleBackPressAction = new DoubleBackPressAction() {
    @Override
    public void actionCall() {
          // TODO : Exiting application code
          finish();
    }
};

// setup DoubleBackPress behaviour
DoubleBackPress doubleBackPress = new DoubleBackPress()
        .withDoublePressDuration(3000)                  // timeout duration - msec
        .withDoubleBackPressAction(doubleBackPressAction);

并在活动中将新行为设置为背压的默认行为。

@Override
public void onBackPressed() {
    doubleBackPress.onBackPressed();
}

Example GIF of similar requirements

Answer 10

与Toast一起使用的简便解决方案

在Java中

private Toast exitToast;

@Override
public void onBackPressed() {
    if (exitToast == null || exitToast.getView() == null || exitToast.getView().getWindowToken() == null) {
        exitToast = Toast.makeText(this, "Press again to exit", Toast.LENGTH_LONG);
        exitToast.show();
    } else {
        exitToast.cancel();
        super.onBackPressed();
    }
}

在科特林

private var exitToast: Toast? = null

override fun onBackPressed() {
    if (exitToast == null || exitToast!!.view == null || exitToast!!.view.windowToken == null) {
        exitToast = Toast.makeText(this, "Press again to exit", Toast.LENGTH_LONG)
        exitToast!!.show()
    } else {
        exitToast!!.cancel()
        super.onBackPressed()
    }
}

Answer 11

定义全局变量名称Counter。

int counter = 0;

在您 MainActivity.java
并按照以下步骤操作

在Java中

    @Override
    public void onBackPressed() {
        counter+=1;
        if (counter==2){
            super.onBackPressed();
        }else {
            Toast.makeText(this, "Press one more time to exit", Toast.LENGTH_SHORT).show();
        }

    }

Answer 12

定义一个布尔变量名称doubleBackToExitPressedOnce。

boolean doubleBackToExitPressedOnce = true;

在MainActivity.java中重写onBackPressed（）方法

并按照以下步骤操作，

@Override
public void onBackPressed() {
    if (drawer.isDrawerOpen(GravityCompat.START)) {
        drawer.closeDrawer(GravityCompat.START);
    } else {
        if (doubleBackToExitPressedOnce) {
            this.doubleBackToExitPressedOnce = false;
            Globals.showToast(this, "Please press back again to exit.");
        } else {
            finish();
        }
    }
}

单击Android后退按钮两次退出应用程序

12 个答案: