Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • Home
  • SEARCH
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8981771
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 15, 20262026-06-15T20:25:09+00:00 2026-06-15T20:25:09+00:00

Hi I’m following the example here and I get it to work. But problem

  • 0

Hi I’m following the example here and I get it to work.
But problem is that I don’t have a list to display so I wonder how do I modify the functions so that they don’t need a list element in the xml to work?

I want my ” R.layout.fragment_pager_list” to look like this

<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>

I.E not having to have the

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

In it. How can I modify

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);
        }
    }
}

In order to achieve this?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-15T20:25:10+00:00Added an answer on June 15, 2026 at 8:25 pm

    If you do not want lists, do not use a ListFragment for the pages. Create your own Fragment with your own page UI. For example, here is a sample project that uses fragments, each hosting an EditText widget.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have a French site that I want to parse, but am running into
I have a .ini file as follows: [playlist] numberofentries=2 File1=http://87.230.82.17:80 Title1=(#1 - 365/1400) Example
I have a small JavaScript validation script that validates inputs based on Regex. I
I'm parsing an RSS feed that has an &#8217; in it. SimpleXML turns this
This could be a duplicate question, but I have no idea what search terms
I don't have much knowledge about the IPv6 protocol, so sorry if the question
I'm trying to convert HTML to plain text. I get many &\#8217; &\#8220; etc.
I have been unable to fix a problem with Java Unicode and encoding. The
I ran into a problem. Wrote the following code snippet: teksti = teksti.Trim() teksti
I have a string like this: La Torre Eiffel paragonata all&#8217;Everest What PHP function

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.