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嗨，我在这里按照示例进行操作，并且可以正常工作。但问题是我没有要显示的列表，所以我想知道如何修改函数以便它们不需要 xml 中的列表元素来工作？

我希望我的“R.layout.fragment_pager_list”看起来像这样

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:orientation="vertical"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:background="@android:drawable/gallery_thumb">

    <TextView android:id="@+id/text"
        android:layout_width="match_parent" android:layout_height="wrap_content"
        android:gravity="center_vertical|center_horizontal"
        android:textAppearance="?android:attr/textAppearanceMedium"
        android:text="@string/hello_world"/>

    <!-- The frame layout is here since we will be showing either
    the empty view or the list view.  -->
    <FrameLayout
        android:layout_width="match_parent"
        android:layout_height="0dip"
        android:layout_weight="1" >

        <!-- Here is the view to show if the list is emtpy -->
        <TextView android:id="@android:id/empty"
            android:layout_width="match_parent"
            android:layout_height="match_parent"
            android:textAppearance="?android:attr/textAppearanceMedium"
            android:text="No items."/>

    </FrameLayout>

</LinearLayout>

IE不必有 

<ListView android:id="@android:id/list"
        android:layout_width="match_parent"
        android:layout_height="match_parent"
        android:drawSelectorOnTop="false"/>

在里面。我该如何修改

public static class ArrayListFragment extends ListFragment {
        int mNum;

        /**
         * Create a new instance of CountingFragment, providing "num"
         * as an argument.
         */
        static ArrayListFragment newInstance(int num) {
            ArrayListFragment f = new ArrayListFragment();

            // Supply num input as an argument.
            Bundle args = new Bundle();
            args.putInt("num", num);
            f.setArguments(args);

            return f;
        }

        /**
         * When creating, retrieve this instance's number from its arguments.
         */
        @Override
        public void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            mNum = getArguments() != null ? getArguments().getInt("num") : 1;
        }

        /**
         * The Fragment's UI is just a simple text view showing its
         * instance number.
         */
        @Override
        public View onCreateView(LayoutInflater inflater, ViewGroup container,
                Bundle savedInstanceState) {
            View v = inflater.inflate(R.layout.fragment_pager_list, container, false);
            View tv = v.findViewById(R.id.text);
            ((TextView)tv).setText("Fragment #" + mNum);
            return v;
        }

        @Override
        public void onActivityCreated(Bundle savedInstanceState) {
            super.onActivityCreated(savedInstanceState);
            setListAdapter(new ArrayAdapter<String>(getActivity(),
                    android.R.layout.simple_list_item_1, Cheeses.sCheeseStrings));
        }

        @Override
        public void onListItemClick(ListView l, View v, int position, long id) {
            Log.i("FragmentList", "Item clicked: " + id);
        }
    }
}

为了达到这个目的？

4

1 回答 1

1

如果您不想要列表，请不要将 aListFragment用于页面。Fragment使用您自己的页面 UI创建您自己的。例如，这是一个使用片段的示例项目，每个片段都托管一个EditText小部件。

于 2012-12-13T14:05:17.310 回答