Jerry Wilcox [EMAIL PROTECTED] wrote:
I'm not sure about the count in a year, but
I frequently need to
determine how many of a given day of the
week fall in a given month
of the year, or, more precisely, given that
today is Saturday,
September 20, I need to figure out whether
today is the
First: How many of a particular DOW are in a
period. This is just a matter of dividing the
total days by seven then adding one if
needed. If a function were added for this,
then I imagine its more a part of
DateTime::Span rather than DateTime itself.
I already said this. :)
Secondly you
At 11:40 PM -0500 9/19/03, Dave Rolsky wrote:
DateTime-day_count_in_year( year = 1999, day = 4 );
I think this is what was originally meant.
What would people need this info for?
I'm not sure about the count in a year, but I frequently need to
determine how many of a given day of the week
I'm not sure about the count in a year, but I frequently need to
determine how many of a given day of the week fall in a given month
of the year, or, more precisely, given that today is Saturday,
September 20, I need to figure out whether today is the first,
second, third, fourth, or fifth
On 9/20/03 Joshua Hoblitt wrote:
I'm not sure about the count in a year, but I frequently need to
determine how many of a given day of the week fall in a given month
of the year, or, more precisely, given that today is Saturday,
September 20, I need to figure out whether today is the first,
Hi Dave Josh,
Dave Rolsky [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Wed, 17 Sep 2003, Hill, Ronald wrote:
[snipped]
Josh is confused because you're passing arguments to the day_of_week
method. These arguments are completely ignored by DateTime.pm!
-dave
Ok, I see I
Syamala Tadigadapa wrote:
Here is a simple solution
... I'd hate to see your complex solution ...
Rick
On Wed, 17 Sep 2003, Ron Hill wrote:
Ok, I see I can just do
my $dow = $dt2-day_of_week(
year = $dt-year,
);
The day_of_week() method DOES NOT TAKE ARGUMENTS!
I don't know what you think the code above does, but I can tell you that
all it does is return the day of the
The day_of_week() method DOES NOT TAKE ARGUMENTS!
I wonder if it's worth the overhead of checking for extraneous parameters on all
methods?
-J
--
On Fri, 19 Sep 2003, Joshua Hoblitt wrote:
The day_of_week() method DOES NOT TAKE ARGUMENTS!
I wonder if it's worth the overhead of checking for extraneous
parameters on all methods?
I'd rather try to keep accessors as quick as possible.
-dave
/*===
House Absolute
Hi Ron,
I'm a bit confused by your parameters to day_of_week().
This is the actual implementation from DateTime.pm
sub day_of_week { $_[0]-{local_c}{day_of_week} }
-J
--
script=
use strict;
use warnings;
use DateTime;
my $dt =
Hi Josh,
Hi Ron,
I'm a bit confused by your parameters to day_of_week().
This is the actual implementation from DateTime.pm
sub day_of_week { $_[0]-{local_c}{day_of_week} }
-J
I compute the number of days in the current year
I figure out the day of week for january the
On Wed, 17 Sep 2003, Hill, Ronald wrote:
I checked the docs for datetime and used them
F:\scriptsperldoc DateTime|grep day_of_week
File STDIN:
$dow= $dt-day_of_week; # 1-7 (Monday is 1) - also dow, wday
_0. So for example, this class provides both day_of_week() and
Here is a simple solution (unless you are bent on doing it in a longer way
using a date time class.)
sub jan1{
my $y = shift;
my $m = 1; $d = 1;
$m = 11; $y--;
my $c = int($y / 100); $yy = $y %100;
my $z = ( 1 + $yy + int($yy/4) + int($c/4) - 2*$c) % 7;
$z += 7 if $z 0;
return $z; #
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