我的活动中有一些碎片
[1], [2], [3], [4], [5], [6]
如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。
最好的做法是什么?
编辑:应用程序不能从[3]…[6]返回[2]
我的活动中有一些碎片
[1], [2], [3], [4], [5], [6]
如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。
最好的做法是什么?
编辑:应用程序不能从[3]…[6]返回[2]
当前回答
我宁愿这样做:
private final static String TAG_FRAGMENT = "TAG_FRAGMENT";
private void showFragment() {
final Myfragment fragment = new MyFragment();
final FragmentTransaction transaction = getSupportFragmentManager().beginTransaction();
transaction.replace(R.id.fragment, fragment, TAG_FRAGMENT);
transaction.addToBackStack(null);
transaction.commit();
}
@Override
public void onBackPressed() {
final Myfragment fragment = (Myfragment) getSupportFragmentManager().findFragmentByTag(TAG_FRAGMENT);
if (fragment.allowBackPressed()) { // and then you define a method allowBackPressed with the logic to allow back pressed or not
super.onBackPressed();
}
}
其他回答
如果您管理将每个事务添加到后退堆栈的流程,那么您可以这样做,以便在用户按后退按钮时显示上一个片段(您也可以映射home按钮)。
@Override
public void onBackPressed() {
if (getFragmentManager().getBackStackEntryCount() > 0)
getFragmentManager().popBackStack();
else
super.onBackPressed();
}
我认为最简单的方法是创建一个接口,并在Activity中检查片段是否属于接口类型,如果是,则调用它的方法来处理弹出。下面是要在片段中实现的接口。
public interface BackPressedFragment {
// Note for this to work, name AND tag must be set anytime the fragment is added to back stack, e.g.
// getActivity().getSupportFragmentManager().beginTransaction()
// .replace(R.id.fragment_container, MyFragment.newInstance(), "MY_FRAG_TAG")
// .addToBackStack("MY_FRAG_TAG")
// .commit();
// This is really an override. Should call popBackStack itself.
void onPopBackStack();
}
下面是如何实现它。
public class MyFragment extends Fragment implements BackPressedFragment
@Override
public void onPopBackStack() {
/* Your code goes here, do anything you want. */
getActivity().getSupportFragmentManager().popBackStack();
}
在你的Activity中,当你处理弹出时(可能在onBackPressed和onOptionsItemSelected中),使用这个方法弹出backstack:
public void popBackStack() {
FragmentManager fm = getSupportFragmentManager();
// Call current fragment's onPopBackStack if it has one.
String fragmentTag = fm.getBackStackEntryAt(fm.getBackStackEntryCount() - 1).getName();
Fragment currentFragment = getSupportFragmentManager().findFragmentByTag(fragmentTag);
if (currentFragment instanceof BackPressedFragment)
((BackPressedFragment)currentFragment).onPopBackStack();
else
fm.popBackStack();
}
最理想的方法如下: 片段:当按下后退按钮并自定义时调用的回调
public class MyActivity extends Activity
{
//...
//Defined in Activity class, so override
@Override
public void onBackPressed()
{
super.onBackPressed();
myFragment.onBackPressed();
}
}
public class MyFragment extends Fragment
{
//Your created method
public static void onBackPressed()
{
//Pop Fragments off backstack and do your other checks
}
}
rootView.setFocusableInTouchMode(true);
rootView.requestFocus();
rootView.setOnKeyListener(new View.OnKeyListener() {
@Override
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK) {
Fragment NameofFragment = new NameofFragment;
FragmentTransaction transaction=getFragmentManager().beginTransaction();
transaction.replace(R.id.frame_container,NameofFragment);
transaction.commit();
return true;
}
return false;
}
});
return rootView;
对于那些谁使用静态片段
在这种情况下,如果你有一个静态片段,那么它会更可取。 为片段创建一个实例对象
private static MyFragment instance=null;
在MyFragment的onCreate()中初始化该实例
instance=this;
也可以创建一个函数来获取Instance
public static MyFragment getInstance(){
return instance;
}
也可以创建函数
public boolean allowBackPressed(){
if(allowBack==true){
return true;
}
return false;
}
//allowBack is a boolean variable that will be set to true at the action
//where you want that your backButton should not close activity. In my case I open
//Navigation Drawer then I set it to true. so when I press backbutton my
//drawer should be get closed
public void performSomeAction(){
//.. Your code
///Here I have closed my drawer
}
在你能做的活动中
@Override
public void onBackPressed() {
if (MyFragment.getInstance().allowBackPressed()) {
MyFragment.getInstance().performSomeAction();
}
else{
super.onBackPressed();
}
}