Question

我有一个日期范围，我希望能够反向循环。给出以下内容，我将如何实现此目标，标准Range运算符似乎无法正常工作。

>> sd = Date.parse('2010-03-01')
=> Mon, 01 Mar 2010
>> ed = Date.parse('2010-03-05')
=> Fri, 05 Mar 2010
>> (sd..ed).to_a
=> [Mon, 01 Mar 2010, Tue, 02 Mar 2010, Wed, 03 Mar 2010, Thu, 04 Mar 2010, Fri, 05 Mar     2010]
>> (ed..sd).to_a
=> []

如您所见，范围运算符从头到尾正常工作，但不是从头到尾。

Answer 1

尝试上/下：

irb(main):003:0> sd = Date.parse('2010-03-01')
=> #<Date: 4910513/2,0,2299161>
irb(main):004:0> ed = Date.parse('2010-03-15')
=> #<Date: 4910541/2,0,2299161>
irb(main):005:0> sd.upto(ed) { |date| puts date }
2010-03-01
2010-03-02
2010-03-03
2010-03-04
2010-03-05
2010-03-06
2010-03-07
2010-03-08
2010-03-09
2010-03-10
2010-03-11
2010-03-12
2010-03-13
2010-03-14
2010-03-15
=> #<Date: 4910513/2,0,2299161>
irb(main):006:0> ed.downto(sd) { |date| puts date }
2010-03-15
2010-03-14
2010-03-13
2010-03-12
2010-03-11
2010-03-10
2010-03-09
2010-03-08
2010-03-07
2010-03-06
2010-03-05
2010-03-04
2010-03-03
2010-03-02
2010-03-01
=> #<Date: 4910541/2,0,2299161>

Answer 2

我通常只是反转生成的数组：

ruby-1.8.7-p72 > sd = Date.parse('2010-03-01')
 => Mon, 01 Mar 2010 
ruby-1.8.7-p72 > ed = Date.parse('2010-03-05')
 => Fri, 05 Mar 2010 
ruby-1.8.7-p72 > (sd..ed).to_a
 => [Mon, 01 Mar 2010, Tue, 02 Mar 2010, Wed, 03 Mar 2010, Thu, 04 Mar 2010, Fri, 05 Mar 2010] 
ruby-1.8.7-p72 > (sd..ed).to_a.reverse
 => [Fri, 05 Mar 2010, Thu, 04 Mar 2010, Wed, 03 Mar 2010, Tue, 02 Mar 2010, Mon, 01 Mar 2010]

我想，当你不知道开始日期是在结束日期之前还是之后，为了让它做正确的事情，你需要的是：

def date_range(sd, ed)
  sd < ed ? (sd..ed).to_a : (ed..sd).to_a.reverse
end

这将为你提供正确的选择：

ruby-1.8.7-p72 > sd = Date.parse('2010-03-01')
 => Mon, 01 Mar 2010 
ruby-1.8.7-p72 > ed = Date.parse('2010-03-05')
 => Fri, 05 Mar 2010 
ruby-1.8.7-p72 > date_range(sd, ed)
 => [Mon, 01 Mar 2010, Tue, 02 Mar 2010, Wed, 03 Mar 2010, Thu, 04 Mar 2010, Fri, 05 Mar 2010] 
ruby-1.8.7-p72 > date_range(ed, sd)
 => [Fri, 05 Mar 2010, Thu, 04 Mar 2010, Wed, 03 Mar 2010, Tue, 02 Mar 2010, Mon, 01 Mar 2010]

如何以反向循环日期范围？

2 个答案: