lua日期处理函数

function day_step(old_day,step)
   local y,m,d
   if("0" ~= string.sub(old_day,6,6)) then
      m=string.sub(old_day,6,7)
   else
      m=string.sub(old_day,7,7)
   end

if("0" ~= string.sub(old_day,9,9)) then
      d=string.sub(old_day,9,10)
   else
      d=string.sub(old_day,10,10)
   end

y=string.sub(old_day,0,4)
   
   local old_time=os.time{year=y,month=m,day=d}
   local new_time=old_time+86400*step

local new_day=os.date("*t",new_time)
   local res=""

if(tonumber(new_day.day)<10 and tonumber(new_day.month)<10)then
      res=new_day.year.."-".."0"..new_day.month.."-".."0"..new_day.day
   elseif tonumber(new_day.month)<10 then
      res=new_day.year.."-".."0"..new_day.month.."-"..new_day.day
   
   elseif tonumber(new_day.day)<10 then
      res=new_day.year.."-"..new_day.month.."-".."0"..new_day.day
   else
      res=new_day.year.."-"..new_day.month.."-"..new_day.day
   end
   return res
end