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:
<syntaxhighlight lang="lua"> local spellnum = require('Module:Spellnum per MOS').spellnum spellnum{ 11 } -- returns 'eleven' </syntaxhighlight>
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 = 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