返回键（基于Fragment）

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目标：基于Activity的返回很好搞，那如何制作基于Fragment的返回键呢？

关键方法：

假设类调用顺序为 A -> B，B 通过返回键回到A，e.g.

A.java (这里是TongueFragment)

public void initTongueResult(Activity activity) {final String TONGUE_RESULT = "tongueResultFragment";FragmentTransaction transaction = activity.getFragmentManager().beginTransaction();// hide掉当前的tongueFragment tongueFragment = activity.getFragmentManager().findFragmentById(R.id.content);if (tongueFragment instanceof TongueFragment) {transaction.hide(tongueFragment);}// 创建或显示tongueResultFragment tongueResultFragment = activity.getFragmentManager().findFragmentByTag(TONGUE_RESULT);if (tongueResultFragment != null && tongueResultFragment instanceof TongueResultFragment) {transaction.show(tongueResultFragment);} else {tongueResultFragment = new TongueResultFragment();transaction.add(R.id.content, tongueResultFragment, TONGUE_RESULT);}transaction.commit();}

B.java (extends Fragment，这里是TongueResultFragment)

uiNavigationHeader.setActivity(getActivity());

public void goBack() {final String TONGUE = "tongueFragment";FragmentTransaction transaction = getActivity().getFragmentManager().beginTransaction();// 干掉当前的tongueResultFragment tongueResultFragment = getActivity().getFragmentManager().findFragmentById(R.id.content);if (tongueResultFragment instanceof TongueResultFragment) {transaction.remove(tongueResultFragment);}// 返回之前的tongueFragment tongueFragment = getActivity().getFragmentManager().findFragmentByTag(TONGUE);if (tongueFragment != null && tongueFragment instanceof TongueFragment) {transaction.show(tongueFragment);}transaction.commit();}

辅助XML：

B对应XML

    <Button        android:id="@+id/btn_left"        android:layout_width="wrap_content"        android:layout_height="wrap_content"        android:layout_marginLeft="10.0dip"        android:background="@drawable/ic_launcher"        android:text="返回"        android:textColor="@android:color/white" />

Manifest

        <activity android:name=".TongueResultFragment">      </activity>

上面的代码还不是最精简的情况，但大体思路基本如此，以作参考，有不明白的地方，欢迎随时联系。

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