Getting number of days in a month

[Shandy Nantz]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have built, the
idea being that when a month turns from February to March, for example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know when
to stop counting. Any ideas, Thanks,

-S

 

[Gregory Seidman]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have built, the
idea being that when a month turns from February to March, for example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know when
to stop counting. Any ideas, Thanks,

require 'date'

def days_in_month(month, year)
month = month.to_i
year = year.to_i
raise ArgumentError.new("invalid month") unless (1..12).to_a.include? month
first = Date.parse sprintf("%04d%02d01", year, month)
next_month = first + 32
(last - last.mday).mday
end

-S

--Greg

 

[Thomas Preymesser]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have built, the
idea being that when a month turns from February to March, for example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know when
to stop counting. Any ideas, Thanks,

i have not completely understood your problem, but there is a gem call
'date_utils' which does various calendar calculations.

See http://thopre.wordpress.com/2007/06/15/ruby-gem-of-the-week-200724/ for
some examples!

-Thomas

--=20
Thomas Preymesser
(e-mail address removed)
(e-mail address removed)
B=FCro: 030 - 830 353 88
mobil: 0176 - 75 03 03 04
Privat: 030 - 49 78 37 06
http://thopre.wordpress.com/
http://www.thopre.com/

 

[Morton Goldberg]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have
built, the
idea being that when a month turns from February to March, for
example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know
when
to stop counting. Any ideas, Thanks,

<code>
require "Date"
d = Date.new(2008, 2, -1)
d.day # => 29
</code>

You can find the last day of a month by creating a Date object for
day -1.

Regards, Morton

 

[Brian Adkins]

require 'date'

def days_in_month(month, year)
month = month.to_i
year = year.to_i
raise ArgumentError.new("invalid month") unless (1..12).to_a.include? month
first = Date.parse sprintf("%04d%02d01", year, month)
next_month = first + 32
(last - last.mday).mday
end


--Greg

might want to try running that before posting

 

[Gregory Seidman]

might want to try running that before posting

Ah, details. Change the last line of the method to:

(next_month + next_month.mday).mday

Anyhow, it's worth noting that ActiveSupport includes Time.days_in_month.

--Greg

 

[Brian Adkins]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have built, the
idea being that when a month turns from February to March, for example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know when
to stop counting. Any ideas, Thanks,

require 'date'

def days_in_month year, month
((month < 12) ?
Date.new(year, month + 1) :
Date.new(year + 1, 1)).-(1).mday
end

puts days_in_month(2008,2) # => 29

 

[Brian Adkins]

<code>
require "Date"
d = Date.new(2008, 2, -1)
d.day # => 29
</code>

You can find the last day of a month by creating a Date object for
day -1.

Regards, Morton

And we have a winner

I wish I had read your post before posting mine. I need to read the
stdlib doc more carefully.

 

[Brian Adkins]

Ah, details. Change the last line of the method to:

(next_month + next_month.mday).mday

Anyhow, it's worth noting that ActiveSupport includes Time.days_in_month.

--Greg

You still didn't run it, did you? <sigh>

Some ideas you may want to consider:
1) it's probably reasonable to expect numeric month and day arguments,
so you can skip the .to_i calls
2) instead of creating a range, converting it to an array and calling
include?, wouldn't it be better to just use a simple comparison such
as "unless month > 0 && month < 13
3) sprintf'ing a date just to parse it is unnecessary & inefficient
4) it's still broken

 

[Gregory Seidman]

You still didn't run it, did you? <sigh>

Don't sigh at me. I wrote some code off the cuff and fired it off. I also
didn't include unit tests. Yes, it was buggy and inefficient, but it got
across the approach I was using.

Some ideas you may want to consider:
1) it's probably reasonable to expect numeric month and day arguments,
so you can skip the .to_i calls

Given that it was for clarity, I think it's valuable.

2) instead of creating a range, converting it to an array and calling
include?, wouldn't it be better to just use a simple comparison such
as "unless month > 0 && month < 13

Arguable. I prefer range inclusion to a pair of comparisons, but that's a
matter of taste. The to_a only matters if I hadn't performed a to_i on the
month argument previously.

3) sprintf'ing a date just to parse it is unnecessary & inefficient

True enough. Date.new (a.k.a. Date.civil) takes year, month, and day
arguments. For that matter, as pointed out elsewhere in this thread, a -1
for the day argument gives the last day of the month, making the rest of
the method moot.

4) it's still broken

Typo. The + should have been a - in the correction. The correct, if
unnecessary, method is:

require 'date'

def days_in_month(month, year)
month = month.to_i
year = year.to_i
raise ArgumentError.new("invalid month") unless (1..12).include? month
first = Date.civil(year, month, 1)
next_month = first + 32
(next_month - next_month.mday).mday
end

An even simpler method, taken from elsewhere in the thread:

require 'date'

def days_in_month(month, year)
Date.civil(year, month, -1).mday
end

--Greg

 

[Dan Fitzpatrick]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have built, the
idea being that when a month turns from February to March, for example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know when
to stop counting. Any ideas, Thanks,

-S

If you don't want to use the Date module, this is the plain old way:

def days_in_month(m=Time.new.month,y=Time.new.year)
return [31,0,31,30,31,30,31,31,30,31,30,31][m-1] unless m == 2
((y % 4 == 0) and ( (!(y % 100 == 0)) or (y % 400 == 0) ) ) ? 29 : 28
end

Dan

 

[Rob Biedenharn]

Don't sigh at me. I wrote some code off the cuff and fired it off. I
also
didn't include unit tests. Yes, it was buggy and inefficient, but it
got
across the approach I was using.

The point is applicable to *everyone*:

"Don't present code that you haven't actually
run unless you *say* that you haven't run it."

You'll just end up confusing the OP or other readers that may benefit
from
the code. If you're trying to illustrate a point or an approach, say so
or make it quite clear that your code is incomplete or intended as
pseudo-code (particularly if it appears to be valid syntax).

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)

 

[7stud --]

Don't sigh at me. I wrote some code off the cuff and fired it off. I
also
didn't include unit tests. Yes, it was buggy and inefficient, but it got
across the approach I was using.

Thanks for caring, however this isn't a creative writing forum.

 

[James Britt]

The point is applicable to *everyone*:

"Don't present code that you haven't actually
run unless you *say* that you haven't run it."

Nah, better to let dubious code pop up now and then so that people are
not lulled into a false sense of security, and will hopefully realize
that, even if someone says they ran and tested the code and assure you
it's fine, the code may still be bad.



--
James Britt

"In physics the truth is rarely perfectly clear, and that is certainly
universally the case in human affairs. Hence, what is not surrounded by
uncertainty cannot be the truth."
- R. Feynman

 

[Todd Benson]

The point is applicable to *everyone*:

"Don't present code that you haven't actually
run unless you *say* that you haven't run it."

You'll just end up confusing the OP or other readers that may benefit
from
the code. If you're trying to illustrate a point or an approach, say so
or make it quite clear that your code is incomplete or intended as
pseudo-code (particularly if it appears to be valid syntax).

-Rob

I frequently write bad code when I use a system where I cannot cut and
paste. All code on a forum, IMHO, should be taken with a grain of
salt. I don't think posted code needs to be perfect.

2c,
Todd

 

[Shandy Nantz]

<code>
require "Date"
d = Date.new(2008, 2, -1)
d.day # => 29
</code>
Regards, Morton

This is what I ened up using and it seems to work. Thanks,

-S

 

[William James]

This is probably an easy question but I am trying to get at the number
of days that are in a month. I have this calendar that I have built, the
idea being that when a month turns from February to March, for example,
the calendar should redisplay itself properly formated showing the new
month and the correct number of days. I have it so that it starts
counting the days on the right day of the week, but I have to know when
to stop counting. Any ideas, Thanks,

-S

Awk (not my code):

function days_in_month( y, m )
{
return m==2 ? 28+!(y%4)-!(y%100)+!(y%400) : 30+(m%2 != (m>7))
}

 

[Serg Koren]

A more interesting question is how do you get the number of weeks in a
year.
Some years have 52...some have 54. Calendar weeks...not 7 day weeks.

S

 

[Rick DeNatale]

A more interesting question is how do you get the number of weeks in a
year.
Some years have 52...some have 54. Calendar weeks...not 7 day weeks.

I assume you mean here how many 7 day weeks starting on a particular
week day have at least one day in a given year.

If this is the case then in all but one case every year has 53
'weeks.' The only exception is a leap year which starts on the last
day of the week which has 54 weeks.

for start in (1..7) do
for days in (365..366) do
s = "a #{{365 => 'normal', 366 => 'leap'}[days]} year starting on
weekday #{start} starts with "
start_week_days = start == 1 ? 0 : (8 - start)
s << "a 'week' of #{start_week_days} days then " unless start_week_days == 0
days_remaining = days - start_week_days
full_weeks = days_remaining / 7
end_week_days = days_remaining % 7
s << "#{full_weeks} full weeks"
s << " and ends with a 'week' of #{end_week_days} days" unless
end_week_days == 0
s << " a total of #{(start_week_days > 0 ? 1 : 0) + full_weeks +
(end_week_days > 0 ? 1 : 0)} 'weeks.'"
puts s
end
puts
end


a normal year starting on weekday 1 starts with 52 full weeks and ends
with a 'week' of 1 days a total of 53 'weeks.'
a leap year starting on weekday 1 starts with 52 full weeks and ends
with a 'week' of 2 days a total of 53 'weeks.'

a normal year starting on weekday 2 starts with a 'week' of 6 days
then 51 full weeks and ends with a 'week' of 2 days a total of 53
'weeks.'
a leap year starting on weekday 2 starts with a 'week' of 6 days then
51 full weeks and ends with a 'week' of 3 days a total of 53 'weeks.'

a normal year starting on weekday 3 starts with a 'week' of 5 days
then 51 full weeks and ends with a 'week' of 3 days a total of 53
'weeks.'
a leap year starting on weekday 3 starts with a 'week' of 5 days then
51 full weeks and ends with a 'week' of 4 days a total of 53 'weeks.'

a normal year starting on weekday 4 starts with a 'week' of 4 days
then 51 full weeks and ends with a 'week' of 4 days a total of 53
'weeks.'
a leap year starting on weekday 4 starts with a 'week' of 4 days then
51 full weeks and ends with a 'week' of 5 days a total of 53 'weeks.'

a normal year starting on weekday 5 starts with a 'week' of 3 days
then 51 full weeks and ends with a 'week' of 5 days a total of 53
'weeks.'
a leap year starting on weekday 5 starts with a 'week' of 3 days then
51 full weeks and ends with a 'week' of 6 days a total of 53 'weeks.'

a normal year starting on weekday 6 starts with a 'week' of 2 days
then 51 full weeks and ends with a 'week' of 6 days a total of 53
'weeks.'
a leap year starting on weekday 6 starts with a 'week' of 2 days then
52 full weeks a total of 53 'weeks.'

a normal year starting on weekday 7 starts with a 'week' of 1 days
then 52 full weeks a total of 53 'weeks.'
a leap year starting on weekday 7 starts with a 'week' of 1 days then
52 full weeks and ends with a 'week' of 1 days a total of 54 'weeks.'

Or did you mean something else.

 

[Serg Koren]

Good job. That's exactly what I meant.

A more interesting question is how do you get the number of weeks
in a
year.
Some years have 52...some have 54. Calendar weeks...not 7 day weeks.

I assume you mean here how many 7 day weeks starting on a particular
week day have at least one day in a given year.

If this is the case then in all but one case every year has 53
'weeks.' The only exception is a leap year which starts on the last
day of the week which has 54 weeks.

for start in (1..7) do
for days in (365..366) do
s = "a #{{365 => 'normal', 366 => 'leap'}[days]} year starting on
weekday #{start} starts with "
start_week_days = start == 1 ? 0 : (8 - start)
s << "a 'week' of #{start_week_days} days then " unless
start_week_days == 0
days_remaining = days - start_week_days
full_weeks = days_remaining / 7
end_week_days = days_remaining % 7
s << "#{full_weeks} full weeks"
s << " and ends with a 'week' of #{end_week_days} days" unless
end_week_days == 0
s << " a total of #{(start_week_days > 0 ? 1 : 0) + full_weeks +
(end_week_days > 0 ? 1 : 0)} 'weeks.'"
puts s
end
puts
end


a normal year starting on weekday 1 starts with 52 full weeks and ends
with a 'week' of 1 days a total of 53 'weeks.'
a leap year starting on weekday 1 starts with 52 full weeks and ends
with a 'week' of 2 days a total of 53 'weeks.'

a normal year starting on weekday 2 starts with a 'week' of 6 days
then 51 full weeks and ends with a 'week' of 2 days a total of 53
'weeks.'
a leap year starting on weekday 2 starts with a 'week' of 6 days then
51 full weeks and ends with a 'week' of 3 days a total of 53 'weeks.'

a normal year starting on weekday 3 starts with a 'week' of 5 days
then 51 full weeks and ends with a 'week' of 3 days a total of 53
'weeks.'
a leap year starting on weekday 3 starts with a 'week' of 5 days then
51 full weeks and ends with a 'week' of 4 days a total of 53 'weeks.'

a normal year starting on weekday 4 starts with a 'week' of 4 days
then 51 full weeks and ends with a 'week' of 4 days a total of 53
'weeks.'
a leap year starting on weekday 4 starts with a 'week' of 4 days then
51 full weeks and ends with a 'week' of 5 days a total of 53 'weeks.'

a normal year starting on weekday 5 starts with a 'week' of 3 days
then 51 full weeks and ends with a 'week' of 5 days a total of 53
'weeks.'
a leap year starting on weekday 5 starts with a 'week' of 3 days then
51 full weeks and ends with a 'week' of 6 days a total of 53 'weeks.'

a normal year starting on weekday 6 starts with a 'week' of 2 days
then 51 full weeks and ends with a 'week' of 6 days a total of 53
'weeks.'
a leap year starting on weekday 6 starts with a 'week' of 2 days then
52 full weeks a total of 53 'weeks.'

a normal year starting on weekday 7 starts with a 'week' of 1 days
then 52 full weeks a total of 53 'weeks.'
a leap year starting on weekday 7 starts with a 'week' of 1 days then
52 full weeks and ends with a 'week' of 1 days a total of 54 'weeks.'

Or did you mean something else.

--
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/