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