如何将270921sec转换为天+小时+分钟+秒？(红宝石)

Question

我有几秒钟.让我们说270921.我怎么能显示这个数字,说它是xx天,yy小时,zz分钟,ww秒？

Answer 1

它可以非常简洁地使用divmod:

t = 270921
mm, ss = t.divmod(60)            #=> [4515, 21]
hh, mm = mm.divmod(60)           #=> [75, 15]
dd, hh = hh.divmod(24)           #=> [3, 3]
puts "%d days, %d hours, %d minutes and %d seconds" % [dd, hh, mm, ss]
#=> 3 days, 3 hours, 15 minutes and 21 seconds

你可以通过创造性地collect或者可能来进一步干燥它inject,但是当核心逻辑是三条线时它可能是过度的.

Answer 2

我希望有一种比使用divmod更简单的方法,但这是我发现的最干燥和可重复使用的方式:

def seconds_to_units(seconds)
  '%d days, %d hours, %d minutes, %d seconds' %
    # the .reverse lets us put the larger units first for readability
    [24,60,60].reverse.inject([seconds]) {|result, unitsize|
      result[0,0] = result.shift.divmod(unitsize)
      result
    }
end

通过更改格式字符串和第一个内联数组(即[24,60,60])可以轻松调整该方法.

增强版

class TieredUnitFormatter
  # if you set this, '%d' must appear as many times as there are units
  attr_accessor :format_string

  def initialize(unit_names=%w(days hours minutes seconds), conversion_factors=[24, 60, 60])
    @unit_names = unit_names
    @factors = conversion_factors

    @format_string = unit_names.map {|name| "%d #{name}" }.join(', ')
    # the .reverse helps us iterate more effectively
    @reversed_factors = @factors.reverse
  end

  # e.g. seconds
  def format(smallest_unit_amount)
    parts = split(smallest_unit_amount)
    @format_string % parts
  end

  def split(smallest_unit_amount)
    # go from smallest to largest unit
    @reversed_factors.inject([smallest_unit_amount]) {|result, unitsize|
      # Remove the most significant item (left side), convert it, then
      # add the 2-element array to the left side of the result.
      result[0,0] = result.shift.divmod(unitsize)
      result
    }
  end
end

例子:

fmt = TieredUnitFormatter.new
fmt.format(270921)  # => "3 days, 3 hours, 15 minutes, 21 seconds"

fmt = TieredUnitFormatter.new(%w(minutes seconds), [60])
fmt.format(5454)  # => "90 minutes, 54 seconds"
fmt.format_string = '%d:%d'
fmt.format(5454)  # => "90:54"

请注意,format_string不会让您更改部件的顺序(它始终是最重要的值).对于更细粒度的控制,您可以split自己使用和操作值.

Answer 3

需要休息一下.高尔夫球:

s = 270921
dhms = [60,60,24].reduce([s]) { |m,o| m.unshift(m.shift.divmod(o)).flatten }
# => [3, 3, 15, 21]

Answer 4

Rails有一个帮助器,用于转换单词中的时间距离.您可以查看其实现:distance_of_time_in_words

Answer 5

如果你正在使用Rails,如果你不需要精度,有一个简单的方法:

time_ago_in_words 270921.seconds.from_now
# => 3 days