Question

我正在尝试找到生成以下输出的最佳方法

<name> job took 30 seconds
<name> job took 1 minute and 20 seconds
<name> job took 30 minutes and 1 second
<name> job took 3 hours and 2 minutes

我开始使用此代码

def time_range_details
  time = (self.created_at..self.updated_at).count
  sync_time = case time 
    when 0..60 then "#{time} secs"       
    else "#{time/60} minunte(s) and #{time-min*60} seconds"
  end
end

有没有更有效的方法来做到这一点。对于一些非常简单的东西，似乎有很多冗余代码。

另一个用途是：

<title> was posted 20 seconds ago
<title> was posted 2 hours ago

此代码类似，但我使用的是Time.now：

def time_since_posted
  time = (self.created_at..Time.now).count
  ...
  ...
end

Answer 1

如果你需要比distance_of_time_in_words更“精确”的东西，你可以写下这些内容：

def humanize secs
  [[60, :seconds], [60, :minutes], [24, :hours], [Float::INFINITY, :days]].map{ |count, name|
    if secs > 0
      secs, n = secs.divmod(count)

      "#{n.to_i} #{name}" unless n.to_i==0
    end
  }.compact.reverse.join(' ')
end

p humanize 1234
#=>"20 minutes 34 seconds"
p humanize 12345
#=>"3 hours 25 minutes 45 seconds"
p humanize 123456
#=>"1 days 10 hours 17 minutes 36 seconds"
p humanize(Time.now - Time.local(2010,11,5))
#=>"4 days 18 hours 24 minutes 7 seconds"

哦，关于你的代码的一句话：

(self.created_at..self.updated_at).count

是真的糟糕的方式来获得差异。使用简单：

self.updated_at - self.created_at

Answer 2

DateHelper中有两种方法可以为您提供所需的内容：

<强> time_ago_in_words

time_ago_in_words( 1234.seconds.from_now ) #=> "21 minutes"

time_ago_in_words( 12345.seconds.ago )     #=> "about 3 hours"

<强> distance_of_time_in_words

distance_of_time_in_words( Time.now, 1234.seconds.from_now ) #=> "21 minutes"

distance_of_time_in_words( Time.now, 12345.seconds.ago )     #=> "about 3 hours"

Answer 3

chronic_duration将数字时间解析为可读，反之亦然

Answer 4

如果你想在几秒到几天的范围内显示重要的持续时间，那么另一种选择就是（因为它没有表现出最佳效果）：

def human_duration(secs, significant_only = true)
  n = secs.round
  parts = [60, 60, 24, 0].map{|d| next n if d.zero?; n, r = n.divmod d; r}.
    reverse.zip(%w(d h m s)).drop_while{|n, u| n.zero? }
  if significant_only
    parts = parts[0..1] # no rounding, sorry
    parts << '0' if parts.empty?
  end
  parts.flatten.join
end
start = Time.now
# perform job
puts "Elapsed time: #{human_duration(Time.now - start)}"

human_duration(0.3) == '0'
human_duration(0.5) == '1s'
human_duration(60) == '1m0s'
human_duration(4200) == '1h10m'
human_duration(3600*24) == '1d0h'
human_duration(3600*24 + 3*60*60) == '1d3h'
human_duration(3600*24 + 3*60*60 + 59*60) == '1d3h' # simple code, doesn't round
human_duration(3600*24 + 3*60*60 + 59*60, false) == '1d3h59m0s'

或者你可能只对无关紧要的秒段部分感兴趣（也展示了另一种方法）：

def human_duration(duration_in_seconds)
  n = duration_in_seconds.round
  parts = []
  [60, 60, 24].each{|d| n, r = n.divmod d; parts << r; break if n.zero?}
  parts << n unless n.zero?
  pairs = parts.reverse.zip(%w(d h m s)[-parts.size..-1])
  pairs.pop if pairs.size > 2 # do not report seconds when irrelevant
  pairs.flatten.join
end

希望有所帮助。

Answer 5

distance_of_time_in_words如果你通过 1小时30分钟就会出现问题>它会返回约2小时

只需添加帮助：

 PERIODS = {
   'day' => 86400,
   'hour' => 3600,
   'minute' => 60
   }


def formatted_time(total)
  return 'now' if total.zero?

  PERIODS.map do |name, span|
    next if span > total
    amount, total = total.divmod(span)
    pluralize(amount, name)
  end.compact.to_sentence
end

基本上只需几秒钟即可传递数据。

Answer 6

Rails具有DateHelper的视图。如果这不是您想要的，那么您可能必须编写自己的内容。

@MladenJablanović给出了带有良好示例代码的答案。但是，如果您不介意继续自定义示例人性化方法，那么这可能是一个很好的起点。

def humanized_array_secs(sec)
  [[60, 'minutes '], [60, 'hours '], [24, 'days ']].inject([[sec, 'seconds']]) do |ary, (count, next_name)|
    div, prev_name = ary.pop

    quot, remain = div.divmod(count)
    ary.push([remain, prev_name])
    ary.push([quot, next_name])
    ary
  end.reverse
end

这为您提供了可以操纵的值和单位名称的数组。

如果第一个元素不为零，则为天数。您可能需要编写代码来处理多天，例如显示几周，几个月和几年。否则，修剪掉前导0的值，然后取下两个。

def humanized_secs(sec)
  return 'now' if 1 > sec

  humanized_array = humanized_array_secs(sec.to_i)
  days = humanized_array[-1][0]
  case
    when 366 <= days
      "#{days / 365} years"
    when 31 <= days
      "#{days / 31} months"
    when 7 <= days
      "#{days / 7} weeks"
    else
      while humanized_array.any? && (0 == humanized_array[-1][0])
        humanized_array.pop
      end
      humanized_array.reverse[0..1].flatten.join
  end
end

该代码甚至可用于红宝石while语句。

如何使用ruby on rails生成人类可读的时间范围

6 个答案: