How to generate a ListView with headers above some sections?

Question

I want to generate a ListView that has some dividers between some of the entries, like it can be seen in some of the property sections. See the example below. I

Answer 1

You can use my SectionedAdapter, if GPLv3 is acceptable (licensed that way due to some upstream code). You can use my MergeAdapter, if you need something more flexible and with a less-limiting license (Apache 2).

Answer 2

I got a solution. I don't know if it is the best one.

I use a custom adapter derived from ArrayAdapter for the list as described in this tutorial. In the adapter class I check if the position in the getView method is a normal row, then I inflate the row layout. If it is the first row from a new group I inflate a headline layout that is a normal row plus the group headline above it.

If you don't want to mix the header into one of your rows. Consider the following solution:

You can overwrite the two methods getItemViewType and getViewTypeCount. You now have a list that can display different rows. You need to check the expected view type for the item in the getView Method and inflate different layouts depending on it.

The list will handle the recycling for you in a way that it will return only correct recycle views to your getView method, this means if the recycleView is not null it can be used to display your current cell.

Answer 3

I think you might be looking for android.widget.ExpandableListView

http://developer.android.com/reference/android/widget/ExpandableListView.html

Answer 4

I'm also interested in an answer to this. There must be a more straightforward way to do this.

In looking at the Adapter, there's a method, Adapter.getItemViewType(int position).

ListView defines a return value, ITEM_VIEW_TYPE_HEADER_OR_FOOTER which indicates if the returned item is a header or footer.

I haven't tried it, but I assume if you create your own Adapter and return an item with the type indicating it is a header or footer, that the ListView will display it appropriately.