java - Android关于双击退出应用的问题

Question

好多程序返回到第一个activity的时候,再按back会弹出吐司提示双击退出程序
在网上查了一下发现都是用keycode来实现的,用onBackPressed能实现同样的效果吗?
两种哪种方式好一点?用java计时器和handle延时发送两种之间哪个好一点?

Answer 1

没那么麻烦，直接用toast的getView().getParent() 判断是不是空就ok了。API 16 测试通过

public class MainActivity extends Activity {


    private Toast toast;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        toast = Toast.makeText(getApplicationContext(), "确定退出？", 0);

    }
    public void onBackPressed() {
        quitToast();
    }
    /*
    public boolean onKeyDown(int keyCode, KeyEvent event) {
        System.out.println(keyCode + "...." + event.getKeyCode());
        if(keyCode == KeyEvent.KEYCODE_BACK){
            quitToast();
        }
        return super.onKeyDown(keyCode, event);
    }
    */
    private void quitToast() {
        if(null == toast.getView().getParent()){
            toast.show();
        }else{
            System.exit(0);
        }
    }
}

Answer 2

onbackpressed可以,http://tianmaying.com/snippet/8ab3eda84dd8bc9f014de5eab9bf036c
话说你确定有好多程序用到了双击退出?
哪个好点不好说 我一般用onbackpressed.

Answer 3

/* 上一次按返回按键的时间 */
long preBackPressTime;
/* 按返回按键的次数 */
long pressTimes;

@Override
public void onBackPressed() {
    super.onBackPressed();
    long cBackPressTime = SystemClock.uptimeMillis();
    if (cBackPressTime - preBackPressTime < 2000) {
        pressTimes++;
        if (pressTimes >= 2) {
            finish();
        }
    } else {
        pressTimes = 1;
    }
    if (pressTimes == 1) {
        Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();
    }
    preBackPressTime = cBackPressTime;
}