Module:Spellnum per MOS/sandbox

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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'

The above documentation is transcluded from Module:Spellnum per MOS/doc. (edit | history)
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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
-- WARNING: zero defaults to blank which is probably not what you want.
function p.spellnum(args)
	local frame = mw.getCurrentFrame()
	local numeral = tonumber(args[1])
	
		-- Always return as digits 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

	local spelled = frame:expandTemplate{ title = 'spellnum', args = {
		numeral, zero = args['zero'], adj = args['adj'], ord = args['ord'], us = args['us']}}
	
	if mw.ustring.find(spelled,'%a+[ %-]%a+[ %-]%a+') then
		if numeral >= 1000000 and numeral <= 1000000000000000 then
			local size = math.min(4, math.floor(math.log10(numeral) / 3))
			numeral = numeral / 1000^size
			return ({"%.2f ", "%.1f ", "%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