Android实现点击两次返回键退出

来源：互联网发布：北风网的大数据怎样编辑：程序博客网时间：2024/06/04 23:23

第一种方法：

// 定义一个变量，来标识是否退出private static boolean isExit = false;Handler mHandler = new Handler() {    @Override    public void handleMessage(Message msg) {        super.handleMessage(msg);        isExit = false;    }};

@Overrideprotected void onCreate(Bundle savedInstanceState) {    super.onCreate(savedInstanceState);

setContentView(R.layout.activity_main);

}

@Overridepublic boolean onKeyDown(int keyCode, KeyEvent event) {    if (keyCode == KeyEvent.KEYCODE_BACK) {        exit();        return false;    }    return super.onKeyDown(keyCode, event);}private void exit() {    if (!isExit) {        isExit = true;        Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();        // 利用handler延迟发送更改状态信息        mHandler.sendEmptyMessageDelayed(0, 2000);    } else {        finish();        System.exit(0);        ActivityCollector.finishAll();    }}

}

第二种方法（根据点击的时间来判定）：

public class MainActivity extends Activity {

private long exitTime = 0;

@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 false;
}
return super.onKeyDown(keyCode, event);
}

public void exit() {
if ((System.currentTimeMillis() - exitTime) > 2000) {
Toast.makeText(getApplicationContext(), "再按一次退出程序",
Toast.LENGTH_SHORT).show();
exitTime = System.currentTimeMillis();
} else {
finish();
System.exit(0);
}
}

}

0 0