由于红宝石的工作日数，如何获得未来的日期？

Question

我想从开始日期计算结束日期。 我有工作日数（所以没有周日和周六）。 这是我的功能：

def ending_date(start_date, number_of_working_days)
  date = 0
  ending_date = start_date
   while date > number_of_working_days
    next if ending_date.sunday? || ending_date.saturday?
     finish = ending_date + 1.day unless date.saturday? or date.sunday?
     date += 1
   end
   finish
end

我得到nil。我不知道为什么。 如果 somedoby 可以帮助我解决这个问题，请:)

Answer 1

也可以直接计算ending_date，无需任何循环。每5个工作日就是一个周末！我将星期日的代码编写为 start_date，并在一周中的其他日子使用start_date.wday 对其进行了修改。星期六是个特例。

def ending_date(number_of_working_days, start_date = Date.today)
  if start_date.wday == 6 # Saturday is a special case
    start_date + number_of_working_days + (number_of_working_days-1)/5*2 + 1
  else
    start_date + number_of_working_days + (number_of_working_days-1+start_date.wday)/5*2
  end
end

我对许多 start_dates 和许多 number_of_working_days 进行了测试，它似乎工作正常。

不过，我找不到删除 if 语句的方法。

编辑：如果您使用ActiveSupport::TimeWithZone 对象，只需在调用ending_date 之前调用to_date：

time = Time.zone.local(2016, 10, 15, 15, 30, 45)
puts time.class                   #=> ActiveSupport::TimeWithZone
puts ending_date(5, time.to_date) #=> 2016-10-21

Answer 2

枚举器似乎可以胜任，它可以为您提供比日期更多的信息：

require 'date'

def future_working_days(start_date = Date.today)
  date = start_date
  Enumerator.new do |yielder|
    loop do
      date += 1 # NOTE: Should start_date be included?
      yielder << date unless date.saturday? || date.sunday?
    end
  end
end

def ending_date(number_of_working_days, start_date = Date.today)
  future_working_days(start_date).find.with_index(1) { |_, i| i == number_of_working_days }
end

# Show the next 10 working days :
puts future_working_days.take(10).map(&:to_s).join('->') # 2016-12-08->2016-12-09->2016-12-12->2016-12-13->2016-12-14->2016-12-15->2016-12-16->2016-12-19->2016-12-20->2016-12-21

# Date after 5 working days :
puts ending_date(5) #=> 2016-12-14

# How many working days until the next Friday the 13th of a leap year
puts future_working_days.find_index{|date| date.friday? && date.day == 13 && Date.leap?(date.year)} #=> 851

注意事项：

• ending_date 的参数已切换为允许 start_date 的默认日期
• future_working_days 是一个无限枚举器。不要尝试future_working_days.to_a！ Lazy 可能会有所帮助。
• 今天应该包含在future_working_days中吗？

Answer 3

经过最少的更改，它应该如下所示：

def ending_date(start_date, number_of_working_days)
  date = 0
  while date < number_of_working_days
    start_date += 1
    unless start_date.saturday? || start_date.sunday?
      finish = start_date
      date += 1
    end
  end
  finish
end

#> ending_date(Date.today, 5)
#=> #<Date: 2016-12-14 ((2457737j,0s,0n),+0s,2299161j)>

但我更喜欢这样写这个方法：

def ending_date(date, number_of_working_days)
  result = []
  while result.size < number_of_working_days
    date += 1
    unless [6,0].include? date.wday
      result << date
    end
  end
  result 
end

# To take all the working dates
#> ending_date(Date.today, 5)
#=> [#<Date: 2016-12-08 ((2457731j,0s,0n),+0s,2299161j)>, #<Date: 2016-12-09 ((2457732j,0s,0n),+0s,2299161j)>, #<Date: 2016-12-12 ((2457735j,0s,0n),+0s,2299161j)>, #<Date: 2016-12-13 ((2457736j,0s,0n),+0s,2299161j)>, #<Date: 2016-12-14 ((2457737j,0s,0n),+0s,2299161j)>]

# To take last working date
#> ending_date(Date.today, 5).last
#=> #<Date: 2016-12-14 ((2457737j,0s,0n),+0s,2299161j)>