Iterate every month with date objects

Question

So I have two ruby Date objects, and I want to iterate them every month. For example if I have Date.new(2008, 12) and Date.new(2009, 3), it would yield me 2008-12, 2009-1, 2009-

Answer 1

Date.new(2014,1,1).upto(Date.today).map {|date| "#{date.to_s[0..-4]}"}.uniq

Will give you a string representation of each month including it's year.

Answer 2

MonthRange.new(date1..date2).each { |month| ... }
MonthRange.new(date1..date2).map { |month| ... }

You can use all the Enumerable methods if you use this iterator class. I make it handle strings too so that it can take form inputs.

# Iterate over months in a range
class MonthRange
  include Enumerable

  def initialize(range)
    @start_date = range.first
    @end_date   = range.last
    @start_date = Date.parse(@start_date) unless @start_date.respond_to? :month
    @end_date   = Date.parse(@end_date) unless @end_date.respond_to? :month
  end

  def each
    current_month = @start_date.beginning_of_month
    while current_month <= @end_date do
      yield current_month
      current_month = (current_month + 1.month).beginning_of_month
    end
  end
end

Answer 3

I have added following method to Date class:

class Date
  def all_months_until to
    from = self
    from, to = to, from if from > to
    m = Date.new from.year, from.month
    result = []
    while m <= to
      result << m
      m >>= 1
    end

    result
  end
end

You use it like:

>> t = Date.today
=> #<Date: 2009-11-12 (4910295/2,0,2299161)>
>> t.all_months_until(t+100)   
=> [#<Date: 2009-11-01 (4910273/2,0,2299161)>, #<Date: 2009-12-01 (4910333/2,0,2299161)>, #<Date: 2010-01-01 (4910395/2,0,2299161)>, #<Date: 2010-02-01 (4910457/2,0,2299161)>]

Ok, so, more rubyish approach IMHO would be something along:

class Month<Date
  def succ
    self >> 1
  end
end

and

>> t = Month.today
=> #<Month: 2009-11-13 (4910297/2,0,2299161)>
>> (t..t+100).to_a
=> [#<Month: 2009-11-13 (4910297/2,0,2299161)>, #<Month: 2009-12-13 (4910357/2,0,2299161)>, #<Month: 2010-01-13 (4910419/2,0,2299161)>, #<Month: 2010-02-13 (4910481/2,0,2299161)>]

But you would need to be careful to use first days of month (or implement such logic in Month)...

Answer 4

Welp, after lurking 15 years nearly this is my first stack overflow answer, I think.

start_date = Date.new(2000,12,15) # day is irrelevant and can be omitted
end_date = Date.new(2001,2,1). #same

(start_date.to_datetime..end_date.to_datetime).map{|d| [d.year, d.month]}.uniq.sort

# returns [[2000,12],[2001,1],[2001,2]]

(start_date.to_datetime..end_date.to_datetime).map{|d| Date.new(d.year, d.month)}.uniq.sort

# returns an array of date objects for the first day of any month in the span

Answer 5

Here is something very Ruby:

first day of each month

(Date.new(2008, 12)..Date.new(2011, 12)).select {|d| d.day == 1}

It will give you an array of the first day for each month within the range.

last day of each month

(Date.new(2008, 12)..Date.new(2012, 01)).select {|d| d.day == 1}.map {|d| d - 1}.drop(1)

Just note that the end date needs to be the month after your end range.

Answer 6

def each_month(date, end_date)
  ret = []
  (ret << date; date += 1.month) while date <= end_date
  ret
end