两次点击返回键退出应用

我的APP首页是一个H5登录页面，用的WebView，登录之后进入个人首页，登录后在个人首页按返回键时会返回到WebView的登录页面，不能直接退出应用，我想登录后在个人首页点击返回键后直接退出应用。个人分析是Activity栈的原因。WebViewActivity在下面，IndexPageActivity在上面。

转载自：http://blog.csdn.net/sunnyfans/article/details/7688092

上面链接中的方法完美解决了我的问题，感谢！

1. 在项目中键一个SysApplication类，内容如下

import java.util.LinkedList;
import java.util.List;
import android.app.Activity;
import android.app.AlertDialog;
import android.app.Application;
import android.content.DialogInterface;
import android.content.Intent;

public class SysApplication extends Application {
private List<Activity> mList = new LinkedList<Activity>();
private static SysApplication instance;

private SysApplication() {
}
public synchronized static SysApplication getInstance() {
if (null == instance) {
instance = new SysApplication();
}
return instance;
}
// add Activity
public void addActivity(Activity activity) {
mList.add(activity);
}

public void exit() {
try {
for (Activity activity : mList) {
if (activity != null)
activity.finish();
}
} catch (Exception e) {
e.printStackTrace();
} finally {
System.exit(0);
}
}
public void onLowMemory() {
super.onLowMemory();
System.gc();
}
}

2. 然后在每个Activity的onCreate()方法中添加

SysApplication.getInstance().addActivity(this);

具体如下：

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
SysApplication.getInstance().addActivity(this);
requestWindowFeature(Window.FEATURE_NO_TITLE);
setContentView(R.layout.activity_webview);
webView = (WebView) findViewById(R.id.webView);
......
Intent intent = getIntent();
String url = intent.getStringExtra("url");
webView.loadUrl(url);

}

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
SysApplication.getInstance().addActivity(this);
requestWindowFeature(Window.FEATURE_NO_TITLE);
setContentView(R.layout.activity_index_page);
initView();
updateView();
}

        在需要退出的地方调用

SysApplication.getInstance().exit();

我在应用中还使用了点击两次退出，

转载自：http://www.open-open.com/lib/view/open1392186781082.html

具体实现如下：

// 定义一个变量，来标识是否退出
private static boolean isExit = false;

private static Handler mHandler = new Handler() {

@Override
public void handleMessage(Message msg) {
super.handleMessage(msg);
isExit = false;
}
};
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK) {
exit();
return true;
}
return super.onKeyDown(keyCode, event);
}

private void exit() {
if (!isExit) {
isExit = true;
Toast.makeText(getApplicationContext(), "再按一次后退键退出程序",
Toast.LENGTH_SHORT).show();
// 利用handler延迟发送更改状态信息
mHandler.sendEmptyMessageDelayed(0, 2000);
} else {
Log.e(TAG, "exit application");
SysApplication.getInstance().exit();

}
}

把另一种退出方式也摘录上，方便查找

package com.example.clickexittest;

import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
import android.view.KeyEvent;
import android.widget.Toast;

public class MainActivity extends Activity {

private static final String TAG = MainActivity.class.getSimpleName();

private long clickTime = 0; //记录第一次点击的时间

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

}

@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK) {
exit();
return true;
}
return super.onKeyDown(keyCode, event);
}

private void exit() {
if ((System.currentTimeMillis() - clickTime) > 2000) {
Toast.makeText(getApplicationContext(), "再按一次后退键退出程序",
Toast.LENGTH_SHORT).show();
clickTime = System.currentTimeMillis();
} else {
Log.e(TAG, "exit application");
this.finish();
//          System.exit(0);
}
}
}