Did you even *try* to look for this answer on your own? Say, in the
documentation for the function you're using, of the module you're
using?
Straight out of perldoc Date::Calc's Date_to_Days() section :
In order to convert the number returned by this function back into a
date, use the function "Add_Delta_Days()" (described further below), as
follows:
$days = Date_to_Days($year,$month,$day);
($year,$month,$day) = Add_Delta_Days(1,1,1, $days - 1);
Please make an attempt to help yourself before asking hundreds of
people around the world to read documentation to you.
GAAAH. What do you have against newlines?
What made you think the Today() function takes a number of days as an
argument? The docs say:
========================================
# ($year,$month,$day) = Today([$gmt]);
<...>
If the optional (boolean) input parameter "$gmt" is given, a "true"
value ("1") will cause "gmtime()" to be used instead of "localtime()",
internally, thus returning Greenwich Mean Time (GMT, or UTC) instead of
local time.
=======================================
Nothing about this function says that it returns anything other than
*today*'s date.
Paul Lalli