http://www.israelnews2.co.il/weather/
this "year 100" problem is most probably the one I fond on my system
too. anyone using the perl timelocal() as directed in the O'Reilly perl
book should be aware:
($sec,$min,$hour,$mday,$month,$year) = localtime(time);
print $year."\n";
you get: 100!
basicly it's a value that was only "useful" as a two digit year in the
20th century, so "$year += 1900" will always be the right year. the
O'Reilly book fails to mention this, but "perldoc -f localtime" will
tell you all about it. Excerpt:
All array elements are numeric, and come straight out of a struct
tm. In particular this means that $mon has the range 0..11 and
$wday has the range 0..6 with sunday as day 0. Also,
$year is the number of years since 1900, that is, year is
123 in year 2023, and _not_ simply the last two digits of the
year. If you assume it is, then you create non-Y2K-compliant
programs--and you wouldn't want to do that, would you?
therefore, your solution is one of three. after you get the year with
($sec,$min,$hour,$mday,$month,$year) = localtime(time);
you can do one of the following:
$year += 1900; # get the real year number
$year = $year % 100; # get the real "years since last beginning of
# the century"
$year=substr($year,-2,2); # get a 2-numeral string representing the
# year, the way the accountant will be happy.
please publish here more tips and glitches if you find them.
--
Ira Abramov ; whois:IA58 ; www.scso.com ; all around Linux enthusiast
Who wants to remember that escape-x-alt-control-left shift-b puts you into
super-edit-debug-compile mode?
(Discussion in comp.os.linux.misc on the intuitiveness of commands, especially
Emacs.)
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