android 点击两次返回键退出程序实现

1、延时Handler实现

private boolean isExit;

/** 程序是否退出 */
private boolean isFinish;
private final int CODE_NOT_EXIT = 0x000;
private Handler handler = new Handler(){

    public void handleMessage(android.os.Message msg) {
        switch (msg.what) {
        case CODE_NOT_EXIT:
            // 程序退出后不再做操作
            if(isFinish) break;
            isExit = false;
            break;

        default:
            break;
        }
    };
};
@Override
protected void onDestroy() {
    super.onDestroy();
    isFinish = true;
}

@Override
public void onBackPressed() {
    if(isExit){
        super.onBackPressed();
    }else{
        // 两秒内再次点击退出
        Toast.makeText(getApplicationContext(), "再按一次后退键退出程序",Toast.LENGTH_SHORT).show();
        isExit = true;
        handler.sendEmptyMessageDelayed(CODE_NOT_EXIT, 2 * 1000);
    }
}

2、计算当前时间实现

//记录第一次点击的时间
private long clickTime = 0; 
@Override
public void onBackPressed() {  
    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 {
        super.onBackPressed();
    }  
}