如何使用ruby on rails生成人类可读的时间范围
我正在尝试找到生成以下输出的最佳方法
job took 30 seconds job took 1 minute and 20 seconds job took 30 minutes and 1 second job took 3 hours and 2 minutes 我启动了这段代码
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。 对于一些超级简单的东西来说，似乎有很多冗余代码。
另一个用途是：
was posted 20 seconds ago <title> was posted 2 hours ago</code> </pre>
<p> 这个代码是类似的，但我使用Time.now： </p>
<pre> <code>def time_since_posted time = (self.created_at..Time.now).count ... ... end</code> </pre>
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<p> 如果你需要比<code>distance_of_time_in_words</code>更“精确”的东西，你可以写下这些内容： </p>
<pre> <code>def humanize secs [[60, :seconds], [60, :minutes], [24, :hours], [1000, :days]].map{ |count, name| if secs > 0 secs, n = secs.divmod(count) "#{n.to_i} #{name}" 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"</code> </pre>
<p> 哦，你的代码有一句话： </p>
<pre> <code>(self.created_at..self.updated_at).count</code> </pre>
<p> 是一个<em>非常</em>糟糕的方式来获得差异。 使用简单： </p>
<pre> <code>self.updated_at - self.created_at</code> </pre>
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<p> <code>DateHelper</code>中有两种方法可以为您提供所需的方法： </p>
<ol>
<li>
<p> <strong>time_ago_in_words</strong> </p>
<pre> <code>time_ago_in_words( 1234.seconds.from_now ) #=> "21 minutes" time_ago_in_words( 12345.seconds.ago ) #=> "about 3 hours"</code> </pre>
</li>
<li>
<p> <strong>distance_of_time_in_words</strong> </p>
<pre> <code>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"</code> </pre>
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<p> chronic_duration将数字时间解析为可读，反之亦然 </p>
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<p> 如果您想在几秒到几天的范围内显示重要的持续时间，则可以选择（因为它不必执行最佳）： </p>
<pre> <code>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(dhms)).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'</code> </pre>
<p> 或者，您可能只对无关紧要的秒段部分感兴趣（同时演示另一种方法）： </p>
<pre> <code>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(dhms)[-parts.size..-1]) pairs.pop if pairs.size > 2 # do not report seconds when irrelevant pairs.flatten.join end</code> </pre>
<p> 希望有所帮助。 </p>
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<p> <code>distance_of_time_in_words</code>存在问题，如果你将在那里通过<strong>1小时30分钟</strong>它将返回<strong>约2小时</strong> </p>
<p> 只需添加帮助： </p>
<pre> <code> 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</code> </pre>
<p> 基本上只需几秒钟即可传递数据。 </p>
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<p> Rails有一个<code>DateHelper</code>用于视图。 如果这不是您想要的，您可能需要自己编写。 </p>
<p> @MladenJablanović有一个很好的示例代码答案。 但是，如果您不介意继续自定义示例人性化方法，这可能是一个很好的起点。 </p>
<pre> <code>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</code> </pre>
<p> 这为您提供了可以操作的值和单元名称数组。 </p>
<p> 如果第一个元素不为零，则为天数。 您可能希望编写代码来处理多天，例如显示周，月和年。 否则，修剪前导<code>0</code>值，然后取下两个。 </p>
<pre> <code>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</code> </pre>
<p> 该代码甚至可以用于ruby <code>while</code>语句。 </p>
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