一年的周数,有点

Question

最好是红宝石,但逻辑足够好......

鉴于本周非标准,我需要一年中的周数.所以,假设您将周定义为星期六 - >星期五.然后,给定一个日期,这个周数(1-52)是什么？

strftime有%U:

> Time.now.strftime('%U')
> => "28"

...但当然假设标准星期日 - >星期六.

Answer 1

使用%W而不是%U,它使用星期一作为一周的第一天.

Time.now.strftime('%W')

Answer 2

class Date
  def sweek
    date = self + 1
    date.cweek
  end
end

# Today is Sunday, 17 July
Date.today.cweek
#=> 28
Date.today.sweek
#=> 29

Answer 3

也许可以这样做；

def week_dates( week_num )
    year = Time.now.year
    week_start = Date.commercial( year, week_num, 1 )
    week_end = Date.commercial( year, week_num, 7 )
    week_start.strftime( "%m/%d/%y" ) + ' - ' + week_end.strftime( 
"%m/%d/%y" )
  end