最近我在许多Android应用和游戏中注意到这种模式:当点击后退按钮“退出”应用时，Toast会出现类似于“请再次点击后退退出”的消息。
我在想，当我越来越频繁地看到它时，这是一个内置的功能，你可以在某个活动中访问它吗?我已经看了很多类的源代码，但我似乎找不到任何关于这一点。
当然，我可以想到一些很容易实现相同功能的方法(最简单的可能是在活动中保留一个布尔值，指示用户是否已经单击过一次…)，但我想知道这里是否已经有一些东西。
编辑:正如@LAS_VEGAS所提到的，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”，如果这有意义的话:)
当前回答
private static final int TIME_DELAY = 2000;
private static long back_pressed;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
}
@Override
public void onBackPressed() {
if (back_pressed + TIME_DELAY > System.currentTimeMillis()) {
super.onBackPressed();
} else {
Toast.makeText(getBaseContext(), "Press once again to exit!",
Toast.LENGTH_SHORT).show();
}
back_pressed = System.currentTimeMillis();
}
其他回答
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)
当HomeActivity包含导航抽屉和双backPressed()函数退出应用程序。 (不要忘记初始化全局变量布尔doubleBackToExitPressedOnce = false;) 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);
}
}
}
在Kotlin的背面按下退出应用程序，你可以使用:
定义一个全局变量:
private var doubleBackToExitPressedOnce = false
覆盖onBackPressed:
override fun onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed()
return
}
doubleBackToExitPressedOnce = true
Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_LONG).show()
Handler().postDelayed({
doubleBackToExitPressedOnce = false;
}, 2000)
}
这个解的独特之处在于它的行为;其中，非双击将显示吐司和成功双击将显示没有吐司，同时关闭应用程序。 唯一的缺点是吐司的显示将有650毫秒的延迟。我相信这是最佳行为的最佳解决方案，因为逻辑表明，如果没有这样的延迟，就不可能有这种行为
//App Closing Vars
private var doubleBackPressedInterval: Long = 650
private var doubleTap = false
private var pressCount = 0
private var timeLimit: Long = 0
override fun onBackPressed() {
pressCount++
if(pressCount == 1) {
timeLimit = System.currentTimeMillis() + doubleBackPressedInterval
if(!doubleTap) {
showExitInstructions()
}
}
if(pressCount == 2) {
if(timeLimit > System.currentTimeMillis()) {
doubleTap = true
super.onBackPressed()
}
else {
showExitInstructions()
}
pressCount = 1
timeLimit = System.currentTimeMillis() + doubleBackPressedInterval
}
}
private fun showExitInstructions() {
Handler().postDelayed({
if(!doubleTap) {
Toast.makeText(this, "Try Agian", Toast.LENGTH_SHORT).show()
}
}, doubleBackPressedInterval)
}
private static final int TIME_INTERVAL = 2000;
private long mBackPressed;
@Override
public void onBackPressed() {
if (mBackPressed + TIME_INTERVAL > System.currentTimeMillis()) {
super.onBackPressed();
Intent intent = new Intent(FirstpageActivity.this,
HomepageActivity.class);
startActivity(intent);
finish();
return;
} else {
Toast.makeText(getBaseContext(),
"Tap back button twice to go Home.", Toast.LENGTH_SHORT)
.show();
mBackPressed = System.currentTimeMillis();
}
}
