Android 删除碎片问题
我有三个片段: A是我的主片段,B是登录片段,成功登录将进入C片段 当我点击后退按钮时,我需要从片段C移动到片段A 我的问题是,当我点击后退按钮时,我仍然从片段A移动到片段B 如何解决我的问题 下面是我的开关片段函数:Android 删除碎片问题,android,android-fragments,Android,Android Fragments,我有三个片段: A是我的主片段,B是登录片段,成功登录将进入C片段 当我点击后退按钮时,我需要从片段C移动到片段A 我的问题是,当我点击后退按钮时,我仍然从片段A移动到片段B 如何解决我的问题 下面是我的开关片段函数: public void switchFragment(Fragment fragment) { FragmentManager manager = getActivity().getSupportFragmentManager(); FragmentTransac
public void switchFragment(Fragment fragment) {
FragmentManager manager = getActivity().getSupportFragmentManager();
FragmentTransaction transaction = manager.beginTransaction();
transaction.replace(R.id.mainFrame, fragment, null);
transaction.addToBackStack(null);
transaction.commit();
}
A片段(新主页)到B片段(登录片段):
switchFragment(LogInFragment.newInstance());
这是我的B片段,它具有值logged
,用于决定是否切换来自C片段的片段
我认为问题一定在这里,当返回到片段并单击“上一步”按钮想要退出应用程序时,我可以看到logcat显示1=>和2=>
String logged = memberData.getUD_MBTYPENAME(); //get the value when login succeed
Log.d(TAG,"1=>"+logged);
//If UD_MBTYPENAME is not null,change to A fragment
if (!TextUtils.isEmpty(logged)) {
Log.d(TAG,"2=>"+logged);
((MainActivity) getActivity()).switchFragment(NewHomepage.newInstance());
}
以下是我关于onKeyDown和switchFragment的主要活动:
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK) {
if (getSupportFragmentManager().getBackStackEntryCount() == 1) {
quickDialog();//It's a alert dialog
return false;
}
}
return super.onKeyDown(keyCode, event);
}
public void switchFragment(Fragment fragment) {
FragmentManager manager = getSupportFragmentManager();
FragmentTransaction transaction = manager.beginTransaction();
transaction.replace(R.id.mainFrame, fragment, null);
transaction.addToBackStack(null);
transaction.commit();
}
如果我不使用transaction.addToBackStack(null),请引用如下更改代码代码>,问题仍然存在,即使删除transaction.addToBackStack(null)代码>
当我返回A时,我必须单击两次返回以显示警报对话框,我不知道在片段中第一次单击返回时发生了什么
if (!TextUtils.isEmpty(logged)){
Log.d(TAG,"2=>"+logged);
hideFragment(LogInFragment.newInstance());
switchFragment(NewHomepage.newInstance());
}
public void switchFragment(Fragment fragment) {
FragmentManager manager = getActivity().getSupportFragmentManager();
FragmentTransaction transaction = manager.beginTransaction();
transaction.replace(R.id.mainFrame, fragment, null);
//remove it will fix my issue , but I have to click back twice to show alert dialog , I don't know what happened click it first time in A fragment.
//transaction.addToBackStack(null);
transaction.commit();
}
public void hideFragment(Fragment fragment) {
FragmentManager manager = getActivity().getSupportFragmentManager();
FragmentTransaction transaction = manager.beginTransaction();
transaction.hide(fragment);
transaction.commit();
}
删除hideFragment
并使用manager.popBackStack()代码>在switchFragment上,问题将得到解决。尝试
如果已经添加了片段,它将显示该片段
使用fragmentTransaction.show
方法重新使用现有片段,即保存的实例
public void switchFragment (Fragment oldFragment, Fragment newFragment, int frameId) {
boolean addFragment = true;
FragmentManager fragmentManager = getFragmentManager ();
String tag = newFragment.getArguments ().getString ("TAG");
Fragment fragment = fragmentManager.findFragmentByTag (tag);
// Check if fragment is already added
if (fragment != null && fragment.isAdded ()) {
addFragment = false;
}
// Hide previous fragment
String oldFragmentTag = oldFragment.getArguments ().getString (BaseFragment.TAG);
if (!tag.equals (oldFragmentTag)) {
FragmentTransaction hideTransaction = fragmentManager.beginTransaction ();
Fragment fragment1 = fragmentManager.findFragmentByTag (oldFragmentTag);
hideTransaction.hide (fragment1);
hideTransaction.commit ();
}
// Add new fragment and show it
FragmentTransaction addTransaction = fragmentManager.beginTransaction ();
if (addFragment) {
addTransaction.add (frameId, newFragment, tag);
addTransaction.addToBackStack (tag);
}
else {
newFragment = fragmentManager.findFragmentByTag (tag);
}
addTransaction.show (newFragment);
addTransaction.commit ();
}
可能重复嘿谢谢你的帮助,我可以从你的链接找到答案。非常感谢。感谢你的代码管理员,那么这是否意味着在我的例子中,当C到A只是从你的代码中隐藏B?它将隐藏B it oldFragment指的是,当你从B切换到另一个片段时,它隐藏B感谢你的响应,但我发现如果我不删除transaction.addToBackStack(null)代码>,即使我使用“隐藏”或“不隐藏”,问题也不会得到解决。我已经更新了这个问题。