i have the date (which i can provide in any format) eg. 12.12.1972
i need seconds of this date from 1970 as 'time' function gives for the current date
or is it possible to subtract two dates or
exactly the situation is to find a date is between a period
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Actually i cant understand properly ur question but to for the time we normally use gmtime(time) or localtime(time)
in both the cases it gives u 9 elements array which u can use as per ur requirement.
like this.
($sec,$min,$hour,$mday,$mo
gmtime(time);
or
($sec,$min,$hour,$mday,$mo
localtime(time);
I think it will help u
use Time::ParseDate;
$seconds_since_jan1_1970 = parsedate("12/12/1972");
#=========================
# Here's one using Date::Manip.
use Date::Manip;
$seconds_since_jan1_1970 = UnixDate("12/12/1972", "%o");
# find out if a date is between two dates
my $date = ParseDate($d);
my $date1 = ParseDate($d1);
my $date2 = ParseDate($d2);
if ($date >= $date1 and $date <= $date2) {
print "$d is in the range.\n";
}
#=========================
Both Time::ParseDate and Date::Manip supports several date formats. Out of the two
Date::Manip is more featureful and flexible but also is pretty heavy and big.
All,
This question seems to be abandoned. As part of the ongoing cleanup effort, I will leave a note in
the Community Support topic area which will recommend to the Moderators that they close this question.
My recommendation will be to award the points to marecs.
PLEASE DO NOT AWARD THE POINTS TO ME.
Experts ===>: please leave your thoughts here. <====
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Answer for Membership
by: marecsPosted on 2001-05-21 at 00:47:41ID: 6102484
use Time::Local;
$seconds = timegm($sec,$min,$hours,$m
$seconds = timelocal($sec,$min,$hours
in your case
$seconds = timegm(0,0,0,12,11,70);
perldoc Time::Local for more details.