在Ruby/Rails中生成两个日期之间的商业周列表

Question

我正在尝试创建一个年周列表（相当于Rails中mySQL的YEARWEEK(date,1))位于两个日期值之间。如果开始日期和结束日期位于同一年，则列表将完美生成。以下是我的代码：

  campaign_start_date = "2013-08-02 06:59:00"
  campaing_end_date = "2013-09-01 06:59:00"
  start_year = DateTime.parse(campaign_start_date).cwyear
  start_week = "%04d%02d" % [start_year, DateTime.parse(campaign_start_date).cweek]
  end_year = DateTime.parse(campaing_end_date).cwyear
  end_week = "%04d%02d" % [end_year, DateTime.parse(campaing_end_date).cweek]
  if start_year == end_year
    (start_week..end_week).each{ |i| result << i }
  else
    # need to build a suitable logic here. to handle the case when duration spans over multiple years. for example started in 01-Nov-14 and ended in 01-May-15
  end
  return result

上面的日期值不会有问题，它会落在if的情况下，我会得到的结果是：

[
  "201331",
  "201332",
  "201332",
  "201333",
  "201334",
  "201335"
]

这也是我想要的。但是，如果我的开始日期和结束日期值是例如：

  campaign_start_date = "2014-07-23 06:59:00"
  campaing_end_date = "2015-03-01 06:59:00"

意味着落在不同的年份，那么它需要与我在if条件下的逻辑不同，因为对于这些日期值（start_week=201430和end_week=201509），if条件在这里不合适，因为它会生成80值，这是错误的，因为这些日期之间的周数不是80。需要帮助开发else案例的逻辑。这可能很容易，但现在我只是厌倦了再深入挖掘。

特别注意：解决方案应关注商业年度和商业周（参见铁轨的.cwyer和.cweek函数）。例如，2016-01-01的年度周将是201553，而不是201601

在这方面的任何帮助都将不胜感激。

Answer 1

感谢那些回复t的人。我终于解决了这样的问题：

  campaign_weeks = []
  campaign_start_date = "2014-07-23 06:59:00" # or any date
  campaing_end_date = "2015-03-01 06:59:00" # or any date
  start_year = DateTime.parse(campaign_start_date).cwyear
  start_cweek_of_the_campaign = "%04d%02d" % [start_year, DateTime.parse(campaign_start_date).cweek]
  end_year = DateTime.parse(campaing_end_date).cwyear
  end_cweek_of_the_campaign = "%04d%02d" % [end_year, DateTime.parse(campaing_end_date).cweek]
  if start_year == end_year
    (start_cweek_of_the_campaign..end_cweek_of_the_campaign).each do |w|
      campaign_weeks << ("%04d%02d" % [start_year, w])
    end
  else
    (start_year..end_year).each do |y|
      first_cweek_number_of_the_year = (y == start_year) ? start_cweek_of_the_campaign : 1
      last_cweek_number_of_the_year = (y == end_year) ? end_cweek_of_the_campaign : DateTime.new(y, 12, 28).cweek
      (first_cweek_number_of_the_year .. last_cweek_number_of_the_year).each do |w|
        campaign_weeks << ("%04d%02d" % [y, w])
      end
    end
  end
  return campaign_weeks

注：12月28日总是在一年中的最后一周。一年中最后一个ISO周是52或53。

参考：http://en.wikipedia.org/wiki/ISO_week_date#Last_week

从这个答案中得到了一些提示：使用Ruby 计算一年中的周数

Answer 2

瓶颈是(start_week..end_week)范围。它显然超过了一百（因为我们是小数）：

 2014xx ⇒ 201452 ⇒ 201453 ⇒ ... ⇒ 201499 ⇒ 201500 ⇒ ...

你可能应该过滤你的范围，比如：

r = (start_week..end_week)
r.to_a.reject { |e| e[-2..-1].to_i > 52 }

根据您计算周数的方式（=基于-或基于1），201500可能也会被过滤：/

r.to_a.select { |e| e[-2..-1].to_i.between? 1, 52 }

Answer 3

试试这个；无论年份是否相同，它都适用于任何一组日期：

campaign_start_date = "2014-07-23 06:59:00"
campaign_end_date = "2015-03-01 06:59:00"
start_date = DateTime.parse(campaign_start_date)
end_date = DateTime.parse(campaign_end_date)
while start_date < end_date
  puts "%04d%02d" % [start_date.cw_year, start_date.cweek]
  start_date = start_date + 7.days
end

Answer 4

讨论有点晚，但以下是我过去得到的两个日期之间的商业周数：

def cweek_diff(start_date, end_date)
  return if end_date < start_date
  cweek_diff = (end_date.cweek - start_date.cweek) + 1
  cwyear_diff = end_date.cwyear - start_date.cwyear
  cyear_diff * 53 + cweek_diff - cwyear_diff
end

它在我的情况下非常有效。希望它能有所帮助；）