ruby-on-rails - 如何回到前一天？

Question

在我的 activeadmin 上，我必须给出每家商店的开门和关门时间。 例如，如果商店从晚上 9 点营业到凌晨 4 点，则记录为同一天。 但是我的算法出了问题

我的算法是:

  def opened?
    today_day = Date.today.wday
    yesterday_day = today_day == 0 ? 6 : (today_day - 1)
    opening = self.openings.where(day: [today_day, yesterday_day]).first
    if opening
      opening_day = (Time.now).day
      if opening.closes_at < opening.opens_at
        opening_day = yesterday_day
      end
      # binding.pry
      today_opens_at = Time.new((Time.now).year, (Time.now).month, opening_day, opening.opens_at.hour, opening.opens_at.min)
      today_closes_at = Time.new((Time.now).year, (Time.now).month, (Time.now).day, opening.closes_at.hour, opening.closes_at.min)
      if today_opens_at < (Time.now) && (Time.now) < today_closes_at
        true
      else
        false
      end
    else
      false
    end
  end

我尝试为“1.day”添加 gem activesupport，但出现错误“参数超出范围” 我没有找到解决方案，你能帮帮我吗？

Answer 1

yerderday_day 必须是一个月中的某一天 (0..30)

     today_day_of_week = Date.today.wday
     yesterday_day_of_week = today_day_of_week == 0 ? 6 : (today_day_of_week - 1)
     opening = self.openings.where(day: [today_day_of_week, yesterday_day_of_week]).first
     yesterday_day = Date.today.prev_day.day

或者只是替换这个 block :

  if opening.closes_at < opening.opens_at
    opening_day = Date.today.prev_day.day
  end