Question

我已将以下代码添加到我的基本活动中，以在我的应用中实现后退功能

@Override
public boolean onKeyDown(int keyCode, KeyEvent event) 
{
    System.out.println("============This Is Back Pressed==============");
    System.out.println("===============getBackStackEntryCount()==========="+getSupportFragmentManager().getBackStackEntryCount());
    if (keyCode == KeyEvent.KEYCODE_BACK)
     {
        if (getSupportFragmentManager().getBackStackEntryCount() == 0)
        {
            this.finish();
            return false;
        }
        else
        {

            getSupportFragmentManager().popBackStack();
            return false;
        }

     }

    for (MyonBackPressed listener : onBackPressedListeners)
    {
        listener.Onback(keyCode,event);
    }
    System.out.println("====>"+onBackPressedListeners);
    return super.onKeyDown(keyCode, event);
}

但按下后退按钮时出现此错误我在may应用程序中使用滑动菜单库

03-27 14:30:15.835: I/System.out(7118): ============This Is Back Pressed==============
03-27 14:30:15.835: I/System.out(7118): ===============getBackStackEntryCount()===========1
03-27 14:30:15.855: I/System.out(7118): i===>1
03-27 14:30:15.875: I/System.out(7118): all event url       http://54.200.110.49/checkplanner/webservice/alleventlist.php?email=test.prismetric@gmail.com&limit=0
 03-27 14:30:15.875: W/dalvikvm(7118): threadid=1: thread exiting with uncaught exception       (group=0x40bab450)
03-27 14:30:15.935: E/AndroidRuntime(7118): FATAL EXCEPTION: main
03-27 14:30:15.935: E/AndroidRuntime(7118): java.lang.IllegalStateException: The specified child       already has a parent. You must call removeView() on the child's parent first.
03-27 14:30:15.935: E/AndroidRuntime(7118):     at android.view.ViewGroup.addViewInner(ViewGroup.java:3486)
03-27 14:30:15.935: E/AndroidRuntime(7118):     at android.view.ViewGroup.addView(ViewGroup.java:3357)
03-27 14:30:15.935: E/AndroidRuntime(7118):     at android.view.ViewGroup.addView(ViewGroup.java:3302)
03-27 14:30:15.935: E/AndroidRuntime(7118):     at android.view.ViewGroup.addView(ViewGroup.java:3278)
03-27 14:30:15.935: E/AndroidRuntime(7118):     at android.support.v4.app.NoSaveStateFrameLayout.wrap(NoSaveStateFrameLayout.java:40)
03-27 14:30:15.935: E/AndroidRuntime(7118):     at android.support.v4.app.FragmentManagerImpl.moveToState(FragmentManager.java:931)
 03-27 14:30:15.935: E/AndroidRuntime(7118):    at android.support.v4.app.FragmentManagerImpl.moveToState(FragmentManager.java:1104)
0 3-27 14:30:15.935: E/AndroidRuntime(7118):    at android.support.v4.app.BackStackRecord.popFromBackStack(BackStackRecord.java:764)
 03-27 14:30:15.935: E/AndroidRuntime(7118):    at android.support.v4.app.FragmentManagerImpl.popBackStackState(FragmentManager.java:1516)
  03-27 14:30:15.935: E/AndroidRuntime(7118):   at android.support.v4.app.FragmentManagerImpl$2.run(FragmentManager.java:479)

Answer 1

当您在当时向活动添加片段时

getSupportFragmentManager()
        .beginTransaction()
        .setTransition(FragmentTransaction.TRANSIT_FRAGMENT_OPEN)
        .addToBackStack("perent")      // this is store fragment in to stack and you can get back to previous fragment 
        .add(R.id.fragment_container, fragmentObj)
        .commit();

如何从android中的当前片段返回到前一个片段

1 个答案: