Just wanted to thank you guys for responding. For my needs the solution
proposed by Venkat worked almost perfectly. I only needed to modify so
that I would get the actual date that the 5th Saturday fell on. My
subroutine looks like
for ($mon = $month; $mon == $month; $time += 86400) {
my ($wday);
($day, $mon, $wday) = (localtime($time))[3,4, 6];
if ($wday == $week_day) {
$wday_count++;
if ($wday_count == 5) {
$mon = $mon +1;
$year += 1900;
if ($day ne 1) {
return("$mon/$day/$year");
}
}
}
-----Original Message-----
From: Shlomi Fish [mailto:shlo...@iglu.org.il]
Sent: Wednesday, December 23, 2009 6:02 AM
To: beginners@perl.org
Cc: Venkat Saranathan; Johnson, Reginald (GTS)
Subject: Re: Saturdays in a month
On Tuesday 22 Dec 2009 17:27:41 Venkat Saranathan wrote:
> You can code a simple time logic to calculate this. Here is a sample
code.
> (Bugs included :))
Yes, indeed. :-) Bugs are included. See below for my comments.
>
> #!/usr/bin/perl
>
> use Time::Local;
> use strict;
>
You should add use warnings;
> sub day_in_a_month {
> my ($week_day, $month, $year ) = @_;
> my ($day, $hour, $min, $sec, $time);
> my ($mon, $wday_count);
>
> $day = 1;
> $hour = $min = $sec = 0;
> $month--;
> $year -= 1900;
>
> $time = timelocal($sec, $min, $hour, $day, $month, $year);
>
> for ($mon = $month; $mon == $month; $time += 86400) {
86400 is a magic number which some people may not be familiar with. You
should
assign it to a constant. Furthermore, it's not a very robust way to go a
day
forward. You should use POSIX::mktime() instead.
> my ($wday);
> ($mon, $wday) = (localtime($time))[4, 6];
> if ($wday == $week_day) {
> $wday_count++;
> }
> }
> return($wday_count);
> }
>
Here is my version that uses DateTime (a non-core module, but highly
recommended) and seems to work. One can do it more efficiently in an
O(1) (=
constant time) manner by futzing with the initial day-of-week in Month
day #1
and the number of days in a month, but I didn't get to it yet. My
version
works from what I tried:
<<<<<<<<<<<<<<<<<
#!/usr/bin/perl
use strict;
use warnings;
use DateTime;
sub day_in_a_month
{
# week_day is:
# 1-7 (Monday is 1) - also dow, wday
# From http://search.cpan.org/dist/DateTime/lib/DateTime.pm
my ($week_day, $month, $year ) = @_;
my $start = DateTime->new( year => $year, month => $month, day =>
1);
my $end = $start->clone();
$end->add( months => 1);
my $running = $start->clone();
my $count = 0;
while (DateTime->compare($end, $running) != 0)
{
if ($running->day_of_week() == $week_day)
{
$count++;
}
}
continue
{
$running->add (days => 1);
}
return $count;
}
print day_in_a_month(@ARGV), "\n";
>>>>>>>>>>>>>>>>>
<<<<<<<<<<
shlomi:~$ perl num_days.pl 7 12 2009
4
shlomi:~$ perl num_days.pl 7 11 2009
5
shlomi:~$ perl num_days.pl 6 11 2009
4
shlomi:~$ perl num_days.pl 6 12 2009
4
>>>>>>>>>>
Regards,
Shlomi Fish
> # 0 - sunday 6- saturday
> print day_in_a_month(6, 12, 2009), "\n";
>
>
> with warm regards,
> Venkat Saranathan
> Gulf Breeze Software.
>
>
> On Tue, Dec 22, 2009 at 9:51 AM, Johnson, Reginald (GTS) <
>
> reggie_john...@ml.com> wrote:
> > Is there a module that I can use that will tell me the number of
> > Saturdays, or any weekday, if I give it the month and year.
> >
> > Reginald Johnson
> >
> > TSM Backup & Restore Services
> >
> > [image: Picture (Device Independent Bitmap)]
> >
> > Jacksonville, FL
> >
> > 904.218.4620
> >
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--
-----------------------------------------------------------------
Shlomi Fish http://www.shlomifish.org/
Parody on "The Fountainhead" - http://shlom.in/towtf
Bzr is slower than Subversion in combination with Sourceforge.
( By: http://dazjorz.com/ )
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