Question

我正在编写这个代码，它按天映射日期并将它们推送到数组中：

rtn_ary = []
(2.weeks.ago.to_date..Date.today).map do |date|
  rtn_ary << {period: date}
end

如果我的预期结果是基于几周而不是这样的天数，我如何编写代码？

[[{:period=>Mon, 28 Sep 2015}, {:period=>Tue, 05 Oct 2015}, {:period=>Tue, 12 Oct 2015}]

Answer 1

一个简单的解决方案可能是迭代7天而不是1天：

rtn_ary = (2.weeks.ago.to_date..Date.today).step(7).map do |date|
  {period: date}
end
=> [{:period=>Mon, 28 Sep 2015}, {:period=>Mon, 05 Oct 2015}]

有关步骤的更多信息：http://ruby-doc.org/core-1.9.3/Range.html#method-i-step

Answer 2

为了得到你想要的东西，你需要一个额外的步骤：

(2.weeks.ago.to_date..Date.today).map do |date|
     rtn_ary << {period: date.to_formatted_s(:long)}
 end

to_formatted_s()方法将日期转换为格式化字符串，并采用一个参数来定义如何将日期格式化为字符串。

date.to_formatted_s(:short)         # => "10 Nov"
date.to_formatted_s(:number)        # => "20071110"
date.to_formatted_s(:long)          # => "November 10, 2007"
date.to_formatted_s(:long_ordinal)  # => "November 10th, 2007"
date.to_formatted_s(:rfc822)        # => "10 Nov 2007"
date.to_formatted_s(:iso8601)       # => "2007-11-10"

查看导轨Documentation

如何按周将日期推送到数组中？

2 个答案: