问题

This question was here for other languages, so let here be one for Ruby.

How do I calculate number of complete years that have passed from a given date? As you probably have guessed, that's to calculate person's age automatically. The closest one is distance_of_time_in_words Rails helper, so the following template

Jack is <%= distance_of_time_in_words (Time.now, Time.local(1950,03,22)) %> old.

yields

Jack is over 59 years old.

But I need more precise function that yields just number. Is there one?

If there exists some kind of Ruby on Rails helper function for this, this is OK, although pure Ruby solution would be better.

Edit: the gist of the question is that a non-approximate solution is needed. At the 2^nd of March Jack should be 59 years old and the next day he should be 60 years old. Leap years and such should be taken into account.

回答1:

Do you want age as people typically understand it, or are you looking for a precise measure of time elapsed? If the former, there is no need to worry about leap years and other complications. You simply need to compute a difference in years and reduce it if the person has not had a birthday yet this year. If the latter, you can convert seconds elapsed into years, as other answers have suggested.

def age_in_completed_years (bd, d)
    # Difference in years, less one if you have not had a birthday this year.
    a = d.year - bd.year
    a = a - 1 if (
         bd.month >  d.month or 
        (bd.month >= d.month and bd.day > d.day)
    )
    a
end

birthdate = Date.new(2000, 12, 15)
today     = Date.new(2009, 12, 14)

puts age_in_completed_years(birthdate, today)

回答2:

require 'date'

def years_since(dt)
    delta = (Date.today - Date.parse(dt)) / 365
    delta.to_i
end

回答3:

I have a gem/plugin called dotiw that has a distance_of_time_in_words_hash that will return a hash like: { :years => 59, :months => 11, :days => 27 }. From that you could work out if it's near a certain limit.

回答4:

An approach that handles leap years

Whenever you're calculating elapsed years since a date, you have to decide how to handle leap year. Here is my approach, which I think is very readable, and is able to take leap years in stride without using any "special case" logic.

def years_completed_since(start_date, end_date)

  if end_date < start_date
    raise ArgumentError.new(
      "End date supplied (#{end_date}) is before start date (#{start_date})"
    )
  end

  years_completed = end_date.year - start_date.year

  unless reached_anniversary_in_year_of(start_date, end_date)
    years_completed -= 1
  end

  years_completed
end

# No special logic required for leap day; its anniversary in a non-leap
# year is considered to have been reached on March 1.
def reached_anniversary_in_year_of(original_date, new_date)
  if new_date.month == original_date.month
    new_date.day >= original_date.day
  else
    new_date.month > original_date.month
  end
end

回答5:

I came up with the following, based on a similar reasoning as @FMc First, compute the diff between today's year and birthday's year. Then, sum it to birthday and check the resulting date: if it's greater than today, decrease diff by 1. To be used in Rails apps as it relies on ActiveSupport's years method

def age(birthday, today)
  diff = today.year - birthday.year
  (birthday + diff.years > today ) ? (diff - 1) : diff
end

回答6:

Same idea as FM but with a simplified if statement. Obviously, you could add a second argument instead of using current time.

def age(birthdate)
  now = DateTime.now
  age = now.year - birthdate.year
  age -= 1 if(now.yday < birthdate.yday)
  age
end

回答7:

withing http://github.com/radar/dotiw

Jack is <%= distance_of_time_in_words (Time.now, Time.local(1950,03,22)) %> old.

produce

Jack is 60 years old

回答8:

you can use the ruby gem adroit-age

It works for leap years also..

age = AdroitAge.find_age("23/01/1990")

Update

require 'adroit-age'

dob =  Date.new(1990,1,23)
or
dob = "23/01/1990".to_date

age = dob.find_age
#=> 23

回答9:

I think this will always work, even for someone with a birthday near a leap day:

require 'date'

def calculate_age(start_date, end_date)
  end_date.year - start_date.year - ((end_date.month > start_date.month || (end_date.month == start_date.month && end_date.day >= start_date.day)) ? 0 : 1)
end

puts calculate_age( Date.strptime('03/02/1968', '%m/%d/%Y'), Date.strptime('03/02/2010', '%m/%d/%Y'))

The calculated age with this method in the example call above is 42, which is correct despite 1968 being a leap year and the birthday being near a leap day.

Plus, this way there is no need to create a local variable.

回答10:

How about this:

def age_in_years(date)
  # Difference in years, less one if you have not had a birthday this year.
  today = Date.today
  age = today.year - date.year
  age = age - 1 if [date.day, date.month, today.year].join('/').to_date > Date.today
end

回答11:

d2.year - d1.year - (d2.month > d1.month || (d2.month == d1.month && d2.day >= d1.day) ? 0 : 1)

回答12:

How about something like:

def years_diff(from_time,to_time)
  (((to_time - from_time).abs)/ (365 * 24 * 60 * 60)).to_i
end

years_diff(Time.now,Time.local(1950,03,22)) #=> 59
years_diff(Time.now,Time.local(2009,03,22)) #=> 0
years_diff(Time.now,Time.local(2008,03,22)) #=> 1

来源：https://stackoverflow.com/questions/1904097/how-to-calculate-how-many-years-passed-since-a-given-date-in-ruby