应用程序在导航抽屉内的片段内按下后退按钮时退出

Question

当我尝试从子片段移动到父片段时，我的应用程序退出（它不会崩溃）。这里 LandingActivity.java 是我调用片段 ChannelGrid.java 的主要活动，它调用片段 GridMain.java。当我按下后退按钮时片段 GridMain 应用程序中的移动设备退出而不是移动到 ChannelGrid.java。我已将 addToBackStack("tag") 添加到片段并尝试使用 onKey()..我也在不同的设备上测试了我的应用程序..我检查了其他解决方案有同样的问题并尝试过，但没有任何效果..

导航抽屉-LandingActivity.java-ChaanelGrid.java(Fragment)-GridMain.java(Fragment)

登陆活动.java

   currentFragment = new ChaanelGrid();

                currentFragment.setArguments(args);
                frgManager = getFragmentManager();

                frgManager.beginTransaction().add(R.id.content_frame, currentFragment).addToBackStack("tag")
                        .commit();

ChaanelGrid.java

Fragment currentFragment = new GridMain();



                    FragmentManager frgManager;
                    frgManager = getFragmentManager();
                    frgManager.beginTransaction().add(R.id.content_frame, currentFragment).addToBackStack("GridMain")
                            .commit()

GridMain.java（我也试过没有 onKey() 方法，但没有奏效）

rootView.setOnKeyListener(new View.OnKeyListener() {


   @Override
        public boolean onKey(View v, int keyCode, KeyEvent event) {
            if (keyCode == KeyEvent.KEYCODE_BACK)
            {
                Fragment currentFragment = new ChaanelGrid();


                FragmentManager frgManager;
                frgManager = getFragmentManager();
                frgManager.beginTransaction().replace(R.id.content_frame, currentFragment)
                        .commit();
                return true;
            }

logcat 详细

   10-31 21:46:57.954  24452-24452/D/ActivityThread? ACT-AM_ON_PAUSE_CALLED ActivityRecord{41eb8b98 token=android.os.BinderProxy@41bb9828 {xyz/xyz..activity.LandingActivity_}}
10-31 21:46:57.971  24452-24452/ D/ActivityThread? ACT-PAUSE_ACTIVITY_FINISHING handled : 0 / android.os.BinderProxy@41bb9828
10-31 21:46:58.007  24452-24452/ V/InputMethodManager? focusOut: android.widget.GridView@41f06f40 mServedView=android.widget.GridView@41f06f40 winFocus=false
10-31 21:46:58.297  24452-24452/ I/SurfaceTextureClient? [0x5143bc58] frames:44, duration:1.002000, fps:43.883736
10-31 21:46:58.350  24452-24452/ D/OpenGLRenderer? Flushing caches (mode 0)
10-31 21:46:58.432  24452-24452/ D/OpenGLRenderer? Flushing caches (mode 0)
10-31 21:46:58.753  24452-24452/ D/OpenGLRenderer? Flushing caches (mode 0)
10-31 21:46:58.754  24452-24452/ D/OpenGLRenderer? Flushing caches (mode 0)
10-31 21:46:58.755  24452-24452/ D/OpenGLRenderer? Flushing caches (mode 2)
10-31 21:46:58.879  24452-24452/ D/ActivityThread? ACT-DESTROY_ACTIVITY handled : 1 / android.os.BinderProxy@41bb9828

我尝试在 LandingActivity.java 中添加以下内容

@Override

public void onBackPressed() {
    FragmentManager frgManager;
    frgManager = getFragmentManager();

    Fragment fragment = fragmentManager.findFragmentByTag("GridMain");
    if (fragment != null) {
        GridMain gridMain = (GridMain) fragment;
        if (!gridMain.onBackPressed()) {
            super.onBackPressed();
        }
    }
    else {
        super.onBackPressed();
    }
}

并在 GridMain.java 中跟随

   protected boolean onBackPressed() {
        FragmentManager frgManager;
        frgManager = getFragmentManager();
        frgManager.popBackStack();
        return true;
    }

使用日志我检查了正在调用 LandingActivity.java 的 onBackPressed() 但仍然是相同的输出..

Answer 1

这就是我通常处理后退按钮的方式：

// "State Machine" variables: to indicate which is the active Fragment.
private static boolean isHelpShown = false;
protected static boolean isInfoShown = false;
protected static boolean isMainShown = false;
private static boolean isViewShown = false;

// Used to switch between Fragments.
private static enum Fragments
{
    MAIN,
    VIEW,
    HELP,
    INFO
}

@Override
public final void onBackPressed()
{
    if (isMainShown)
    {
        // We're in the MAIN Fragment.
        finish();
    }
    else
    {
        // We're somewhere else, reload the MAIN Fragment.
        showFragment(Fragments.MAIN);
    }
}

private final void showFragment(final Fragments FragmentType)
{
    isViewShown = false;
    isHelpShown = false;
    isInfoShown = false;
    isMainShown = false;

    final FragmentTransaction ft =
        getSupportFragmentManager().beginTransaction();

    /*
    Replace whatever is in the fragment_container view with this
    fragment, and add the transaction to the back stack so the user can
    navigate back.
    */
    switch(FragmentType)
    {
        case HELP:
        {
            ft.replace(R.id.frgMaster, new FRG_Help());
            isHelpShown = true;
            break;
        }
        case INFO:
        {
            ft.replace(R.id.frgMaster, new FRG_Info());
            isInfoShown = true;
            break;
        }
        case VIEW:
        {
            ft.replace(R.id.frgMaster, new FRG_View());
            isViewShown = true;
            break;
        }
        case MAIN:
        {
            ft.replace(R.id.frgMaster, new FRG_Main());
            isMainShown = true;
            break;
        }
    }

    ft.setTransition(FragmentTransaction.TRANSIT_FRAGMENT_OPEN);

    // Finalize the transaction...
    ft.commit();

    getSupportFragmentManager().executePendingTransactions();
}

正如您所看到的，对于我的需求，我仅在我们位于主片段（我退出）或不在主片段（我返回主片段：再次按后退键并退出）时处理。

您可以在 onBackPressed 方法中添加更多检查，以检查您是否位于另一个 Fragment 中，并相应地加载另一个 Fragment。