listview and fragments

Question

The thing is i'm creating a honeycomb app and i can't switch between fragments . Here's my main code :

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    list = (ListView)findViewById(R.id.list);
    A = new FragmentA();
    B = new FragmentB();
    adapter = new ArrayAdapter<String>(this, android.R.layout.simple_list_item_1, items);
    list.setAdapter(adapter);
    list.setOnItemClickListener(mylistener);
}

public void changeFragment(Fragment f){
    FragmentTransaction ft = getFragmentManager().beginTransaction();
    ft.replace(R.id.fragment, f);
    ft.setTransition(FragmentTransaction.TRANSIT_FRAGMENT_FADE);
    ft.addToBackStack(null);
    ft.commit();
}

   public OnItemClickListener mylistener = new OnItemClickListener() {

    @Override
    public void onItemClick(AdapterView<?> parent, View view, int position,
            long id) {
        switch(position){
        case 0 : changeFragment(A); 
            break;
        case 1 : changeFragment(B); 
            break;
        }

    }
};

I know before i put a "replace" i have to remove but how can i remove the one i'm already in ? What is telling me in the log : The specified child already has a parent. You must call removeView() on the child's parent first .

Could you clarify that the fragmentA IS displayed on selecting the first entry in the list? If so what happens when you select the second entry in the list? I didn't quite understand your comment 'how can I remove the one I'm already in?' — PJL, Jun 25 '11 at 16:05
the one where i am now is FragmentA, clicking on "one" will open/re-open the fragmentA, and "two" will open fragmentB, i just want to pass from FragmentA to FragmentB by clicking on items in the list . — Tsunaze, Jun 25 '11 at 19:49
Have a look at the API demos FragmentLayout sample as it does something similar to what you want I believe. Firstly add some protection so when when you click on the same list entry you don't reload the same fragment. Secondly be careful when maintaining references, see [link](http://stackoverflow.com/questions/6250580/fragment-already-added-illegalstateexception) — PJL, Jun 25 '11 at 23:28

score 1 · Accepted Answer · edited May 23 '17 at 12:19

Have a look at the API demos FragmentLayout sample as it does something similar to what you want I believe. There's nothing wrong with calling replace to load the first fragment as that sample demonstrates.

Firstly add some protection so when when you click on the same list entry you don't reload the same fragment.

Secondly be careful when maintaining references to fragments, see link. You may want to recreate the fragments each time or use remove and add instead of replace.

score 0 · Answer 2 · answered Jun 25 '11 at 19:20

0

The problem is not with removing the fragment. The replace method removes the existing one.

I guess the problem is accessing the fragment inside the listener. Try creating a new fragment inside the listener itself. Please post the complete code.

answered Jun 25 '11 at 19:20

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This was my complete code . Do you want me to post FragmentA, FragmentB, and all the XML too ? – Tsunaze Jun 25 '11 at 19:55
There's no problem with accessing the fragment inside the listener. – PJL Jun 25 '11 at 23:22

listview and fragments

2 Answers2