Module:Spellnum per MOS

{{Module rating }}

Usage

Implements Template:Spellnum per MOS.

{{#invoke:Spellnum per MOS|main|number to format|if second arg equals 1, prefer numerals}}

For use by other Lua modules:

local spellnum = require('Module:Spellnum per MOS').spellnum
spellnum{ 11 } -- returns 'eleven'

---

local p = {}
local words = {"thousand", "million", "billion", "trillion"} -- We don't need to go higher than this, no-one knows what an octillion is.

-- For use by other scripts. Takes arguments:
-- - 1: string or number, value to convert
-- - forcenum: string for Template:Yesno, forces a result in digits for all n ≥ 10.
-- - formating options for spellnum: zero, adj, ord, us
function p.spellnum(args)
	local frame = mw.getCurrentFrame()
	local numeral = tonumber(args[1])
	
   local pass_zero = "zero"
    if args['zero'] ~= nil and args['zero'] ~= '' then
    	pass_zero = args['zero']
    end
    
	-- Always return numerals for negative numbers, non-integers, and if (forcenum and numeral >= 10).
	if numeral < 0 or
			math.fmod(numeral, 1) ~= 0 or
			(numeral >= 10 and frame:expandTemplate{ title = 'yesno', args = {args['forcenum']} } == 'yes') then
		return mw.language.getContentLanguage():formatNum(numeral)
	end

	-- Convert numeral to words
	local spelled = 0
	-- frame:expandTemplate{ title = 'spellnum', args = {
	--numeral, zero = pass_zero, adj = args['adj'], ord = args['ord'], us = args['us']}}
	
	-- Return numerals if more than two words would be needed, else return words
	if mw.ustring.find(spelled,'%a+[ %-]%a+[ %-]%a+') then
		-- Handle numbers larger than one million
		if numeral >= 1000000 and numeral <= 1000000000000000 then
			local size = math.min(4, math.floor(math.log10(numeral) / 3))
			numeral = numeral / 1000^size
			return ({"%.1f ", "%d ", "%d "})[1 + math.floor(math.log10(numeral))]:format(numeral) .. words[size]
		end
		return mw.language.getContentLanguage():formatNum(numeral)
	else 
		return spelled
	end
end

function p.main(frame)
	return p.spellnum(frame.args)
end

return p