Ruby：给出一个日期，找到下一个第二或第四个星期二

Question

I can't seem to find an elegant way to do this...

Given a date how can I find the next Tuesday that is either the 2nd or the 4th Tuesday of the calendar month?

For example:

Given 2012-10-19 then return 2012-10-23

or

Given 2012-10-31 then return 2012-11-13

      October               November        
Su Mo Tu We Th Fr Sa    Su Mo Tu We Th Fr Sa  
    1  2  3  4  5  6                 1  2  3  
 7  8  9 10 11 12 13     4  5  6  7  8  9 10 
14 15 16 17 18 19 20    11 12 13 14 15 16 17  
21 22 23 24 25 26 27    18 19 20 21 22 23 24 
28 29 30 31             25 26 27 28 29 30     

Answer 1

Scroll to the bottom if you just want to see what the end result can look like..

Using code snippets from some date processing work I've done recently in ruby 1.9.3.

Some upgrades to DateTime :

require 'date'

class DateTime

  ALL_DAYS = [ 'sunday', 'monday', 'tuesday',
               'wednesday', 'thursday', 'friday', 'saturday' ]

  def next_week
    self + (7 - self.wday)
  end

  def next_wday (n)
    n > self.wday ? self + (n - self.wday) : self.next_week.next_day(n)
  end

  def nth_wday (n, i)
    current = self.next_wday(n)
    while (i > 0)
      current = current.next_wday(n)
      i = i - 1
    end
    current
  end

  def first_of_month
    self - self.mday + 1
  end

  def last_of_month
    self.first_of_month.next_month - 1
  end

end

method_missing Tricks:

I have also supplemented the class with some method missing tricks to map calls from next_tuesday to next_wday(2) and nth_tuesday(2) to nth_wday(2, 2)`, which makes the next snippet easier on the eyes.

class DateTime

  # ...

  def method_missing (sym, *args, &block)
    day = sym.to_s.gsub(/^(next|nth)_(?<day>[a-zA-Z]+)$/i, '\k<day>')
    dindex = ALL_DAYS.include?(day) ? ALL_DAYS.index(day.downcase) : nil
    if (sym =~ /^next_[a-zA-Z]+$/i) && dindex
      self.send(:next_wday, dindex)
    elsif (sym =~ /^nth_[a-zA-Z]+$/i) && dindex
      self.send(:nth_wday, dindex, args[0])
    else
      super(sym, *args, &block)
    end
  end

  def respond_to? (sym)
    day = sym.to_s.gsub(/^(next|nth)_(?<day>[a-zA-Z]+)$/i, '\k<day>')
    (((sym =~ /^next_[a-zA-Z]+$/i) || (sym =~ /^nth_[a-zA-Z]+$/i)) && ALL_DAYS.include?(day)) || super(sym)
  end

end

Example:

Given a date:

today = DateTime.now
second_tuesday = (today.first_of_month - 1).nth_tuesday(2)
fourth_tuesday = (today.first_of_month - 1).nth_tuesday(4)

if today == second_tuesday
  puts "Today is the second tuesday of this month!"
elsif today == fourth_tuesday
  puts "Today is the fourth tuesday of this month!"
else
  puts "Today is not interesting."
end

You could also edit method_missing to handle calls such as :second_tuesday_of_this_month , :fourth_tuesday_of_this_month , etc. I'll post the code here if I decide to write it at a later date.

Answer 2

Take a look at Chronic or Tickle , both are gems for parsing complex times and dates. Tickle in particular will parse recurring times (I think it uses Chronic as well).

Answer 3

Check out this gem, you might be able to figure out your answer

https://github.com/mojombo/chronic/

Answer 4

Since you already use Rails, you don't need the includes, but this works in pure Ruby as well for reference.

require 'rubygems'
require 'active_support/core_ext'

d = DateTime.parse('2012-10-19')
result = nil
valid_weeks = [d.beginning_of_month.cweek + 1, d.beginning_of_month.cweek + 3]
if valid_weeks.include?(d.next_week(:tuesday).cweek)
  result = d.next_week(:tuesday)
else
  result = d.next_week.next_week(:tuesday)
end

puts result

Answer 5

I think you should probably use a library if you're needing to branch out into more interesting logic, but if what you've described is all you need, the code below should help

SECONDS_PER_DAY = 60 * 60 * 24
def find_tuesday_datenight(now)
  tuesdays = [*-31..62].map { |i| now + (SECONDS_PER_DAY * i) }
    .select { |d| d.tuesday? }
    .group_by { |d| d.month }
  [tuesdays[now.month][1], tuesdays[now.month][-2], tuesdays[(now.month + 1) % 12][1]]
    .find {|d| d.yday > now.yday }
end

Loop through the last month and next month, grab the tuesdays, group by month, take the 2nd and the 2nd last tuesday of the current month (If you actually do want the 4th tuesday, just change -2 to 3) and the 2nd tuesday of the next month and then choose the first one after the provided date.

Here's some tests, 4 tuesdays in month, 5 tuesdays in month, random, and your examples:

[[2013, 5, 1], [2013, 12, 1], [2012, 10, 1], [2012, 10, 19], [2012, 10, 31]].each do |t|
  puts "#{t} => #{find_tuesday_datenight(Time.new *t)}"
end

which produces

[2013, 5, 1] => 2013-05-14 00:00:00 +0800
[2013, 12, 1] => 2013-12-10 00:00:00 +0800
[2012, 10, 1] => 2012-10-09 00:00:00 +0800
[2012, 10, 19] => 2012-10-23 00:00:00 +0800
[2012, 10, 31] => 2012-11-13 00:00:00 +0800

I'm sure it could be simplified, and I'd love to hear some suggestions :) (way too late &tired to even bother figuring out what the actual range should be for valid dates ie smaller than -31..62)

Answer 6

so here is the code that will resolve a weekday for a given week in a month (what you asked for with little sugar). You should not have problems if you are running inside rails framework. Otherwise make sure you have active_support gem installed. Method name is stupid so feel free to change it :)

usage: get_next_day_of_week(some_date, :friday, 1)

require 'rubygems'
require 'active_support/core_ext'

def get_next_day_of_week(date, day_name, count)
  next_date = date + (-date.days_to_week_start(day_name.to_sym) % 7)
  while (next_date.mday / 7) != count - 1 do
    next_date = next_date + 7
  end
  next_date
end 

Answer 7

I use the following to calculate Microsoft's patch Tuesday date. It was adapted from some C# code.

require 'date'

#find nth iteration of given day (day specified in 'weekday' variable)
findnthday = 2
#Ruby wday number (days are numbered 0-7 beginning with Sunday)
weekday = 2

today = Time.now
todayM = today.month
todayY = today.year
StrtMonth = DateTime.new(todayY,todayM ,1)

while StrtMonth.wday != weekday do

    StrtMonth = StrtMonth + 1;
end

PatchTuesday = StrtMonth + (7 * (findnthday - 1))