List Adapter setting - How can I not to repeat the list item layout resource?

Question

I have made a ListView with an ArrayAdapter. It works. But I had to put the resource id for the item layout twice: in the adapter definition and in the getView, in the case

Answer 1

You can override ArrayAdapter constructor and call super :

public ArrayAdapter<T>(Context context, TheDataType data) {
   super(context, R.layout.show_row, data);
 }

And store the id in a ArrayAdapter member variable. It avoid the adapter "user" to know what is the view needed for the adapter.

Or you can use a BaseAdapter.

Answer 2

I had looked into the source code of the ArrayAdapter, and it already does all this stuff about view creation:

// citation from public class ArrayAdapter<T>
private View createViewFromResource(int position, View convertView, ViewGroup parent,
        int resource) {
    View view;
    TextView text;
    if (convertView == null) {
        view = mInflater.inflate(resource, parent, false);
    } else {
        view = convertView;
    }

So, we should simply use it:

lvShows.setAdapter(new ArrayAdapter<TvShow>(this, 
                                            R.layout.show_row, 
                                            R.id.nameField, 
                                            allKnownShows){
public View getView(int position, final View convertView, ViewGroup parent) {
    LinearLayout showView = (LinearLayout) super.getView(position,  convertView, parent);

    setAnythingForItem(showView);

    return showView;
}

Attention: we have changed the constructor!

ArrayAdapter allows to use no-TextView item layout. But if it is not TextView itself, it should have one and the id of this very inner TextView field should be given to the constructor as the third parameter. ArrayAdapter needs it to be set, to write there Strings if the array connected to it has String elements.

It wants a TextView always, even if it doesn't need it really, if the array consists of Objects, not Strings. Otherway it checks the item layout for being a TextView and if it is not, throws an error.