create listview programmatically with no xml layout

Question

I found this example to create a custom listview:

public class UsersListActivity extends ListActivity{    
        @Override
        public void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);         
            String[] statesList = {"listItem 1", "listItem 2", "listItem 3"};
            setListAdapter(new ArrayAdapter<String>(this, R.layout.list_item,
                    statesList)); 
            ListView lv = getListView(); 

            lv.setOnItemClickListener(new OnItemClickListener() {
                public void onItemClick(AdapterView<?> parent, View view,
                        int position, long id) {


                     Toast.makeText(getApplicationContext(),
                     "You selected : "+((TextView) view).getText(), Toast.LENGTH_SHORT).show();
     }
            });

        }

}

However, they are using a xml layout for each row. Is this good practice? What if I wanted to create my own layout programmatically and utilize it in the adapter?

I guess it would be easier just to use the xml layout, but it would be nice to know

RajaReddy PolamReddy · Accepted Answer · 2011-10-27 04:12:20Z

1

change adapter like this

setListAdapter(new ArrayAdapter(this, 
                    android.R.layout.simple_expandable_list_item_1, statesList));

answered Oct 27, 2011 at 4:12

RajaReddy PolamReddy

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1 Comment

Adam Over a year ago

well this: R.layout.list_item is pointing to an xml specifically for each row. What if I wanted a method or a new class to handle the layout? How would I target it in my adapter?

Dinesh Prajapati · Accepted Answer · 2011-10-27 04:15:40Z

0

This is the best practice to use the default android layout, if you do not want your customized view. and if you wanted to use your customized list you can always inflate it from xml and render the data according to list item.

answered Oct 27, 2011 at 4:15

Dinesh Prajapati

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