ruby Rails 3.2.8 - 如何从 Rails 获取周数？
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Rails 3.2.8 - How do I get the week number from Rails?
提问by Chim Kan
I would like to know how to get the current week number from Rails and how do I manipulate it:
我想知道如何从 Rails 获取当前周数以及如何操作它：
- Translate the week number into date.
- Make an interval based on week number.
- 将周数转换为日期。
- 根据周数进行间隔。
Thanks.
谢谢。
回答by meagar
Use strftime:
使用strftime：
%U- Week number of the year. The week starts with Sunday. (00..53)%W- Week number of the year. The week starts with Monday. (00..53)
%U- 一年中的周数。一周从星期日开始。(00..53)%W- 一年中的周数。一周从星期一开始。(00..53)
Time.now.strftime("%U").to_i # 43
# Or...
Date.today.strftime("%U").to_i # 43
If you want to add43 weeks (or days,years,minutes, etc...) to a date, you can use 43.weeks, provided by ActiveSupport:
如果你想添加43周（或天数，年，分钟等）的日期，你可以使用43.weeks，通过提供的ActiveSupport：
irb(main):001:0> 43.weeks
=> 301 days
irb(main):002:0> Date.today + 43.weeks
=> Thu, 22 Aug 2013
irb(main):003:0> Date.today + 10.days
=> Sun, 04 Nov 2012
irb(main):004:0> Date.today + 1.years # or 1.year
=> Fri, 25 Oct 2013
irb(main):005:0> Date.today + 5.months
=> Mon, 25 Mar 2013
回答by Mike
You are going to want to stay away from strftime("%U")and "%W".
你会想要远离strftime("%U")和"%W"。
Instead, use Date.cweek.
相反，使用Date.cweek.
The problem is, if you ever want to take a week number and convert it to a date, strftimewon't give you a value that you can pass back to Date.commercial.
问题是，如果您想获取周数并将其转换为日期，strftime则不会为您提供可以传递回Date.commercial.
Date.commercialexpects a range of values that are 1 based.
Date.strftime("%U|%W")returns a value that is 0 based. You would think you could just +1 it and it would be fine. The problem will hit you at the end of a year when there are 53 weeks. (Like what just happened...)
Date.commercial期望一系列基于 1 的值。
Date.strftime("%U|%W")返回一个基于 0 的值。你会认为你可以+1它就可以了。这个问题会在有 53 周的年末出现。（就像刚刚发生的事情......）
For example, let's look at the end of Dec 2015 and the results from your two options for getting a week number:
例如，让我们看看 2015 年 12 月的结束日期以及您获取周数的两个选项的结果：
Date.parse("2015-12-31").strftime("%W") = 52
Date.parse("2015-12-31").cweek = 53
Now, let's look at converting that week number to a date...
现在，让我们看看将周数转换为日期...
Date.commercial(2015, 52, 1) = Mon, 21 Dec 2015
Date.commercial(2015, 53, 1) = Mon, 28 Dec 2015
If you blindly just +1 the value you pass to Date.commercial, you'll end up with an invalid date in other situations:
如果您盲目地将传递给的值 +1，则Date.commercial在其他情况下最终会得到无效日期：
For example, December 2014:
例如，2014 年 12 月：
Date.commercial(2014, 53, 1) = ArgumentError: invalid date
If you ever have to convert that week number back to a date, the only surefire way is to use Date.cweek.
如果您必须将该周数转换回日期，唯一可靠的方法是使用Date.cweek.
回答by Hitham S. AlQadheeb
date.commercial([cwyear=-4712[, cweek=1[, cwday=1[, start=Date::ITALY]]]]) → date
Creates a date object denoting the given week date.
The week and the day of week should be a negative
or a positive number (as a relative week/day from the end of year/week when negative).
They should not be zero.
For the interval
对于区间
require 'date'
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
puts week_dates(22)
EG: Input (Week Number): 22
Output: 06/12/08 - 06/19/08
EG：输入（周数）：22
输出：06/12/08 - 06/19/08
credit: Siep Korteling http://www.ruby-forum.com/topic/125140
信用：Siep Korteling http://www.ruby-forum.com/topic/125140
回答by Henrik N
Date#cweekseems to get the ISO-8601 week number (a Monday-based week) like %Vin strftime (mentioned by @Robban in a comment).
Date#cweek似乎像%Vstrftime一样获得 ISO-8601 周数（基于星期一的周）（@Robban 在评论中提到）。
For example, the Monday and the Sunday of the week I'm writing this:
例如，我在写这个星期的星期一和星期日：
[ Date.new(2015, 7, 13), Date.new(2015, 7, 19) ].map { |date|
date.strftime("U: %U - W: %W - V: %V - cweek: #{date.cweek}")
}
# => ["U: 28 - W: 28 - V: 29 - cweek: 29", "U: 29 - W: 28 - V: 29 - cweek: 29"]
