Too complex for me

I need help from a ruby master… it’s too hard with my present
understanding, any starting key will be apprciated …

from a period range [start_date, end_date] in ayear , I need to
produce a list of special days like the following :
[“2006-10-12”,“2006-11-08”] => { 10 : [ 12 , 31], 11 : [ 1, 8] }
or
[“2006-10-12”,“2006-12-08”] => { 10 : [ 12 , 31], 11 : [ 1, 30] , 12
: [ 1, 12] }
which means, for each month in the range
the month number : [ period covered in the month] ,

where ‘period covered in the month’ is
[ 1st day of the period or 1st day of the month , last day of the
period or last day of the month ]

coudl I get any clue ?

thanks a lot

joss

On 10/18/06, Josselin [email protected] wrote:

    the month number : [ period covered in the month] ,

where ‘period covered in the month’ is
[ 1st day of the period or 1st day of the month , last day of the
period or last day of the month ]

coudl I get any clue ?

The first step, I would say, should be to convert your values from
strings into real Date objects.
require ‘date’
d1 = Date.new 2006, 10, 12
d2 = Date.new 2006, 12, 8

date_range = (d1…d2)
date_range.each do |date|
puts date
end

You can then collect up the months and days into whatever format
you’re looking for.

On 10/19/06, Josselin [email protected] wrote:

    the month number : [ period covered in the month] ,

I had to make a change to the return values. They are now an array (to
maintain order), with the first element being a string looking like
“2006-10”, as the code works for multiple dates, the second element is
an array with the period covered.

Lots of special case code for handling year boundaries. I’m sure this
can be tightened up, I’d be interested to hear views as to how…

require ‘date’

def coverage rg
from = Date.parse(rg[0])
to = Date.parse(rg[1])

cover = []

build a list of all month indexes we need to iterate over

if from.year==to.year
# trivial case, both dates fall within the same year
month_range = (from.month…to.month )
else
# complex case - dates in different years

# first year - form month to december
month_range = (from.month..12).to_a

# more than one year covered?
(to.year-from.year-1).times do |year|
  # jan to dec
  month_range += (1..12).to_a
end
# last year - jan to to.month
month_range += (1..to.month).to_a

end

y = from.year

month_range.each do |m|
if from < Date.new(y,m,1)
first_day = 1
else
first_day = from.day
end

# passing -1 as the day gets the last day of the month
end_of_month = Date.new(y,m,-1)

if to > end_of_month
  last_day = end_of_month.day
else
  last_day = to.day
end

cover << ["#{y}-#{m}",[first_day, last_day]]

# next year?
y +=1 if m==12

end
cover
end

p coverage([“2006-10-12”,“2008-02-12”])

Cheers,
Max

A possible solution…

It clearly shows every step in the process. Usually, the 5
lines within the loop are reduced to 1 or 2 lines, but you’ll
end up with less readable code.
Not a good idea for newbies… ;]

Notice that it isn’t very efficient when handling very long
periods.

gegroet,
Erik V. - http://www.erikveen.dds.nl/


require “date”

start = Date.parse(ARGV.shift)
finish = Date.parse(ARGV.shift)
hash = {}

(start…finish).each do |date|
month =
days = hash[month] || []
first_day = days.shift || date.day
last_day = date.day
hash[month] = [first_day, last_day]
end

p start.to_s
p finish.to_s
p hash

On 2006-10-19 00:37:43 +0200, “Max M.” [email protected] said:

which means, for each month in the range
joss
require ‘date’
month_range = (from.month…to.month )
end
first_day = from.day

Cheers,
Max

Thanks a lot I’ll work on it today…

joss

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