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
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!!
其他回答
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。
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!!
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();
}
}
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)
}
}
Zefnus使用System.currentTimeMillis()的答案是最好的(+1)。我的方法并没有比这更好，但仍然发布它来补充上面的想法。
如果后退按钮按下时吐司不可见，则显示吐司，反之，如果它可见(后退已经在最后一个吐司中按了一次。LENGTH_SHORT time)，然后退出。
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();
}
}
