我怎样才能使这个条件复杂的实例方法更符合Ruby的习惯？

Question

  def short_remaining_time
    difference  = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')

    # To display the short remaining time in an auction listing.
    if difference[:day] == 0 and difference[:hour] >= 1
      "#{difference[:minute]} minutos"
    elsif difference[:day] == 0 and difference[:hour] >= 23
      "#{difference[:hour]} horas"
    else
      if difference[:day] != 1
        "#{difference[:day]} dias"
      else
        "#{difference[:day]} dia"
      end
    end
  end

This method is inside my auction.rb model in my Rails application.

In one of my views, I am listing all auctions in the system, and I also display how much time is remaining before the auction closes.

Depending on the amount of time, I either show the days hours or minutes .

The code is working fine, just looks and feels very clunky. Is there a way to spruce this up a bit?

Answer 1

You can simplify it as below. Note that your code is redundant. If difference[:hour] >= 23 , then that entails difference[:hour] >= 1 , and will be captured by the latter, so the former condition will never be evaluated to true. So that part can be removed.

def short_remaining_time
  difference  = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')
  case day = difference[:day]
  when 0
    if difference[:hour] >= 1 then "#{difference[:minute]} minutos"
    else "#{day} dias"
    end
  when 1 then "#{day} dia"
  else "#{day} dias"
  end
end

Answer 2

I assume you got your inequalities unintentionally not quite right (you need <= not >= ). Also, if you assume that the hours in the difference will always be no more than 23 , you don't need that check (ie, we're assuming the time difference is "normalized"). So I'd modify it this way to keep your original intent:

  def short_remaining_time
    difference  = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')

    # To display the short remaining time in an auction listing.
    if difference[:day] == 0
      if difference[:hour] <= 1
        "#{difference[:minute]} minutos"
      else
        "#{difference[:hour]} horas"
      end
    else
      "#{difference[:day]} dia" + ((difference[:day] == 1) ? "" : "s")
    end
  end

Answer 3

What about

def short_remaining_time
  difference      = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')
  diff_in_minutes = difference[:day] * 3600 + difference[:hour] * 60

  case diff_in_minutes
    when 0..60      then  "#{difference[:minute]} minutos"
    when 61..3600   then  "#{difference[:hour]  } horas"
    when 3600..7200 then  "#{difference[:day]   } dia"
    else                  "#{difference[:day]   } dias"
  end
end