android中碎片的交易结果为空白屏幕

Question

如果有帮助的话，我想要的与本谷歌教程中所做的类似

DiceTable diceTable = new DiceTable();
Logger logger = new Logger();
fragmentTransaction.add(diceTable, DICETABLE_TAG);
fragmentTransaction.add(logger, LOGGER_TAG);
fragmentTransaction.add(R.id.fragment_container, logger);
fragmentTransaction.add(R.id.fragment_container, diceTable);

    Logger logger = (Logger)fragmentManager.findFragmentByTag(LOGGER_TAG);
    DiceTable diceTable = (DiceTable)fragmentManager.findFragmentByTag(DICETABLE_TAG);

    if (diceTable.isVisible()) {
        fragmentTransaction.replace(R.id.fragment_container, logger);

        fragmentTransaction.commit();
        fragmentTransaction.hide(diceTable);
        fragmentTransaction.show(logger);
    }
    else if (logger.isVisible()) {
        fragmentTransaction.replace(R.id.fragment_container, diceTable);

        fragmentTransaction.commit();
        fragmentTransaction.hide(logger);
        fragmentTransaction.show(diceTable);
    }

我不应该这样做吗？

替换碎片时的空白屏幕

Answer 1

尝试用那种方式初始化片段：

private void initFragments() {
    mDiceTable = new DiceTable();
    mLogger = new Logger();
    isDiceTableVisible = true;

    FragmentManager fm = getSupportFragmentManager();
    FragmentTransaction ft = fm.beginTransaction();
    ft.add(R.id.fragment_container, mDiceTable);
    ft.add(R.id.fragment_container, mLogger);
    ft.hide(mLogger);
    ft.commit();
}

然后以那种方式在它们之间翻转：

 private void flipFragments() {
        FragmentManager fm = getSupportFragmentManager();
        FragmentTransaction ft = fm.beginTransaction();
        if (isDiceTableVisible) {
            ft.hide(mDiceTable);
            ft.show(mLogger);
        } else {
            ft.hide(mLogger);
            ft.show(mDiceTable);
        }
        ft.commit();
        isDiceTableVisible = !isDiceTableVisible;
    }