吐司的最佳解决方案

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Kotlin

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()
    }
}

根据正确的答案和评论中的建议，我创建了一个演示，工作绝对很好，并在使用后删除处理程序回调。

MainActivity.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);
        }
    }

}

希望对大家有所帮助!!

你甚至可以让它更简单，不使用hander，只这样做=)

Long firstClick = 1L;
Long secondClick = 0L;

@Override
public void onBackPressed() {
secondClick = System.currentTimeMillis();
    if ((secondClick - firstClick) / 1000 < 2) {
          super.onBackPressed();
    } else {
          firstClick = System.currentTimeMillis();
          Toast.makeText(MainActivity.this, "click BACK again to exit", Toast.LENGTH_SHORT).show();
        }
 }

Sudheesh B Nair对这个问题有一个很好的(被接受的)答案，我认为应该有一个更好的选择，比如;

测量所经过的时间并检查自上次回按以来TIME_INTERVAL毫秒(例如2000)是否经过了什么错误?下面的示例代码使用System.currentTimeMillis();onBackPressed()被调用来存储时间;

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();
}

回到公认的答案批评;使用一个标志来指示它是否在最后的TIME_INTERVAL(比如2000)毫秒内被按下，而set - reset是通过Handler的postDelayed()方法来执行的，这是我想到的第一件事。但是postDelayed()操作应该在活动关闭时取消，删除Runnable。

为了移除Runnable，它不能被声明为匿名，并且必须与Handler一起声明为成员。然后可以适当地调用Handler的removeCallbacks()方法。

下面的示例是演示;

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;为了防止在应用程序关闭后出现toast消息，toast可以被声明为成员，比如mExitToast，并且可以通过mExitToast.cancel()取消;就在super.onBackPressed()之前;调用。

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();
            }
        }
    }

}

使用到事件onBackPressed

@Override
public void onBackPressed() {
   ExitApp.now(this,"Press again for close");
}

或ExitApp.now(这个,R.string.double_back_pressed)

对于需要关闭的更改秒，指定毫秒

ExitApp.now (R.string.double_back_pressed, 5000)

按键2次返回

public void click(View view){
    if (isBackActivated) {
        this.finish();
    }
    if (!isBackActivated) {
        isBackActivated = true;
        Toast.makeText(getApplicationContext(), "Again", Toast.LENGTH_SHORT).show();
        Handler handler = new Handler();
        handler.postDelayed(new Runnable() {
            @Override
            public void run() {
                isBackActivated = false;  // setting isBackActivated after 2 second
            }
        }, 2000);
    }

}