Android: How to put ListView in XML?

Question

I want to make a simple app that shows a list of contacts (name, surname). My code:

package lista.android;

import java.util.*;

import android.app.ListActivity;
import android.os.Bundle;
import android.widget.*;

class Contact{
    String nume;
    String prenume;

    Contact(String nume, String prenume){
        this.nume=nume;
        this.prenume=prenume;
    }

    public String toString(){
        return prenume +" "+ nume;
    }
}

public class Lista1Activity extends ListActivity {
    /** Called when the activity is first created. */
    ArrayList <Contact> lista;
    ArrayAdapter <Contact> adaptor;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        setListAdapter(adaptor);

        lista=new ArrayList<Contact>();
        adaptor=new ArrayAdapter<Contact>(this, R.id.element, lista);

        adaugaContact("Florian", "Iancu");
        adaugaContact("Ioana", "Constantina");

    }

    public void adaugaContact(String nume, String prenume){
        lista.add(new Contact(nume, prenume));
        adaptor.notifyDataSetChanged();
    }
}

In the XML I have the LinearLayout and a TextView that is the list element. When I run it, the simulator says "Sorry, the application [...] has stopped unexpectedly. Please try again." The LogCat tells me I have to have a ListView whose id is android.R.id.lista. If I create a random ListView field in the XML file and give it the "lista" id, it still doesn't work. How to I call that ListView from XML to match something in my Java code? In other words, I know I'm wrong, but where and how do I fix it?

You are trying to set List Adapter first and then creating it. Don't you think this is the wrong!! — Paresh Mayani
– Paresh Mayani, Commented Nov 21, 2011 at 9:42
You need to understand the concept of creating listview first i think. — Paresh Mayani
– Paresh Mayani, Commented Nov 21, 2011 at 9:42

Kannan Suresh · Accepted Answer · 2011-11-21 11:32:23Z

17

In the LinearLayout you have to specify the ListView component.

This is how your main.xml should be:

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:orientation="vertical" >

    <ListView
        android:id="@+id/sampleListView"
        android:layout_width="fill_parent"
        android:layout_height="fill_parent"
        android:layout_marginBottom="10dp"
        android:layout_marginTop="5dp"
        android:background="@android:color/transparent"
        android:cacheColorHint="@android:color/transparent"
        android:divider="#CCCCCC"
        android:dividerHeight="1dp"
        android:paddingLeft="2dp" >
    </ListView>

</LinearLayout>

Then for the listview make a layout: list_item.xml

<?xml version="1.0" encoding="utf-8"?>
<TextView xmlns:android="http://schemas.android.com/apk/res/android"    android:layout_width="fill_parent"    android:layout_height="fill_parent"    android:padding="10dp"    android:textSize="16sp" >
</TextView>

Change your activity class like this:

public class Lista1Activity extends ListActivity {
    ArrayList<Contact> lista;
    ArrayAdapter<Contact> adaptor;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    lista = new ArrayList<Contact>();
    adaugaContact("Florian", "Iancu");
    adaugaContact("Ioana", "Constantina");

     adaptor = new ArrayAdapter<String>(this, R.layout.list_item, lista);

    setListAdapter(adaptor);
    }

    public void adaugaContact(String nume, String prenume) {
    lista.add(new Contact(nume, prenume));
    adaptor.notifyDataSetChanged();
    }
}

edited Nov 21, 2011 at 11:32

answered Nov 21, 2011 at 9:45

Kannan Suresh

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5 Comments

FloIancu Over a year ago

As I stated, I have tried adding a listview field in my XML, but it has to be tied to my Java code. Where do I tie to it and how do I do it? Thanks!

FloIancu Over a year ago

Ok, so you suggest instantiating "adaptor" before using it. It's a good point, but it still doesn't work, same exception: "need to have a listview whose id is android.R.id.list". The documentation says something about ListView lv = getListView(); How do I get that one into my code?

Kannan Suresh Over a year ago

I have edited the answer. Since you have extended a ListActivity, ListView lv = getListView(); would instantiate the ListView in your activity.

Kannan Suresh Over a year ago

You can use list item designs provided by the platform instead of defining your own layout file (list_item.xml ) for the ListAdapter. For example, try using android.R.layout.simple_list_item_1 instead of R.layout.list_item.

FloIancu Over a year ago

And I also should have overwritten the onListItemClick() function. Thanks for your time and answers! :D

Raghav Chopra · Accepted Answer · 2011-11-21 09:59:27Z

1

<?xml version="1.0" encoding="utf-8"?>

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical" >
<ListView
    android:id="@+id/listview"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:divider="#000000"
    android:dividerHeight=".5dpdp"
    android:paddingLeft="1dp" />
</LinearLayout>

answered Nov 21, 2011 at 9:59

Raghav Chopra

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1 Comment

FloIancu Over a year ago

As I stated, I have tried adding a listview field in my XML, but it has to be tied to my Java code. Where do I tie to it and how do I do it? Thanks!

Jan Dragsbaek · Accepted Answer · 2011-11-21 09:39:21Z

-3

It sounds like you need to take a look at the documentation again.

It is very well described there.

answered Nov 21, 2011 at 9:39

Jan Dragsbaek

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