fix(list): fix list display area overflow.

[tbistr]

The number of Items per a page depend on pagination height. However, pagination height may changes if it enabled or disabled. Re-runing updatePagination() resolves this if needed.

Fix #405

[tbistr]

I found a corner-case.
If Items decreased and items fit in one screen height, list splits into 2 pages.
(Seccond page includes only one item.)

I dont have good solution now.

[tbistr]

Support for page increases/decreases.

[tbistr]

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[naglis]

Thank you for having a go at this.

[naglis]

Having the values hardcoded leaks paginationView() implementation details and is brittle in at least two ways:

• It may break if paginationView() is changed in the future.
• It may break if the user changes the PaginationStyle or ArabicPagination styles (e.g. extra top or bottom margin is added).

Calculating the available height here seems to be quite the chicken and egg conundrum and I do not have better suggestions ATM 🤯 🤷

FWIW, it might be helpful to add some tests while fixing this problem to have more certainty about all possible cases in the future. I've used something like this to test locally, maybe it will be useful:

func prepareListModel(m *Model) {
	m.SetShowFilter(false)
	m.SetShowStatusBar(false)
	m.SetShowTitle(false)
	m.SetShowHelp(false)
	m.SetShowPagination(true)
	// Remove the padding on the left.
	m.Styles.TitleBar.UnsetPadding()
	// Use arabic pagination for simpler output.
	m.Paginator.Type = paginator.Arabic
}

func handleCmd(m Model, cmd tea.Cmd) Model {
	for cmd != nil {
		m, cmd = m.Update(cmd)
	}

	return m
}

func expectView(t *testing.T, m Model, expected string) {
	t.Helper()

	actual := m.View()
	if expected != actual {
		t.Fatalf("Error: expected %q, got %q", expected, actual)
	}
}

func TestUpdatePagination(t *testing.T) {
	list := New([]Item{item("a"), item("b"), item("c")}, itemDelegate{}, 10, 4)
	prepareListModel(&list)

	expectView(t, list, "TODO")

	cmd := list.InsertItem(len(list.Items()), item("d"))
	list = handleCmd(list, cmd)

	expectView(t, list, "TODO")
}

[tbistr]

Thank you for your review!
Indeed, I assumed that paginationView() always returns "" in the case of pagination disabled for the future.

Yes, I'm struggling with that very point.

Calculating the available height here seems to be quite the chicken and egg conundrum

Now I'm thinking that resetting paginator property for each time runs of updatePagination() is not bad idea.
This culculates paginationDefaultHeight each time.
Performance will not be that different.

func (m *Model) updatePagination() {
	index := m.Index()
	availHeight := m.height

	// Reset paginator
	m.Paginator.Page = 0
	m.Paginator.PerPage = len(m.VisibleItems())
	m.Paginator.SetTotalPages(len(m.VisibleItems()))
	paginationDefaultHeight := lipgloss.Height(m.paginationView())

	if m.showTitle || (m.showFilter && m.filteringEnabled) {
		availHeight -= lipgloss.Height(m.titleView())
	}
	if m.showStatusBar {
		availHeight -= lipgloss.Height(m.statusView())
	}
	if m.showPagination {
		availHeight -= paginationDefaultHeight
	}
	if m.showHelp {
		availHeight -= lipgloss.Height(m.helpView())
	}

	updatePages := func(availHeight int) {
		m.Paginator.PerPage = max(1, availHeight/(m.delegate.Height()+m.delegate.Spacing()))

		if pages := len(m.VisibleItems()); pages < 1 {
			m.Paginator.SetTotalPages(1)
		} else {
			m.Paginator.SetTotalPages(pages)
		}
	}

	updatePages(availHeight)

	// If pagination is needed, recompute items alignment
	paginationHeight := lipgloss.Height(m.paginationView())
	if m.showPagination && paginationDefaultHeight != paginationHeight {
		availHeight += paginationDefaultHeight
		availHeight -= paginationHeight
		updatePages(availHeight)
	}

	// Restore index
	m.Paginator.Page = index / m.Paginator.PerPage
	m.cursor = index % m.Paginator.PerPage

	// Make sure the page stays in bounds
	if m.Paginator.Page >= m.Paginator.TotalPages-1 {
		m.Paginator.Page = max(0, m.Paginator.TotalPages-1)
	}
}