Question

我怎样才能说出从X年到Y年的周六和周日的所有日期并将它们存储为数组？伪代码将是

(year_today..next_year).get_all_dates_for_saturday_and_sunday

或许也有宝石迎合这个？

Answer 1

试试这个：

(Date.today..Date.today.next_year).select { |date|
  date.sunday? or date.saturday?
}

#=> [Sat, 03 Sep 2016,Sun, 04 Sep 2016,Sat, 10 Sep 2016,Sun, 11 Sep 2016...

Answer 2

(Date.today..(Date.today + 1.year)).select do |date| 
  date.saturday? || date.sunday? 
end # => [Sat, 03 Sep 2016, Sun, 04 Sep 2016, Sat, 10 Sep 2016, ...

这将为您提供104个元素的数组，其中包含每个日期，即当天和今天之间的星期六或星期日。

Answer 3

以下方法强调效率而不是简洁，避免需要确定某一范围内的每一天是否是一周中的某一天（或两天中的一天）。

代码的

require 'date'

def dates_by_years_and_wday(start_year, end_year, wday)
  (first_date_by_year_and_wday(start_year, wday)...
   first_date_by_year_and_wday(end_year+1, wday)).step(7).to_a
end

def first_date_by_year_and_wday(year, wday)
   d = Date.new(year)
   d + (wday >= d.wday ? wday - d.wday : 7 + wday - d.wday)
end

请注意，范围是用三个点定义的，这意味着end_year中的第一个日期被排除。

实施例

SATURDAY = 6
SUNDAY = 0
start_year, end_year = 2015, 2017

dates_by_years_and_wday(start_year, end_year, SATURDAY)
  #=> [#<Date: 2015-01-03 ((2457026j,0s,0n),+0s,2299161j)>,
  #    #<Date: 2015-01-10 ((2457033j,0s,0n),+0s,2299161j)>,
  #    ...   
  #    #<Date: 2017-12-30 ((2458118j,0s,0n),+0s,2299161j)>]
dates_by_years_and_wday(start_year, end_year, SATURDAY).size
  #=> 157

dates_by_years_and_wday(start_year, end_year, SUNDAY)
  #=> [#<Date: 2015-01-04 ((2457027j,0s,0n),+0s,2299161j)>,
  #    #<Date: 2015-01-11 ((2457034j,0s,0n),+0s,2299161j)>,
  #    ...   
  #    #<Date: 2017-12-31 ((2458119j,0s,0n),+0s,2299161j)>]    
dates_by_years_and_wday(start_year, end_year, SUNDAY).size
  #=> 157

如何获得所有特定日期＆＃39;在一个日期范围内

3 个答案: