你想要人们通常理解的年龄,还是你在寻找一个精确的时间度量?如果是前者,就不必担心闰年等并发症。如果这个人今年还没有过生日,你只需要计算年内的差额,并将其减少。如果是后者,您可以将经过的秒数转换为年数,正如其他答案所建议的那样。

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)

require 'date'

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

我有一个gem/plugin叫做 dotiw 那有 distance_of_time_in_words_hash 它将返回一个散列,比如: { :years => 59, :months => 11, :days => 27 } . 如果接近某个极限,你就可以从中得出结论。

处理闰年的方法

每当你计算一个日期之后经过的年份时,你必须决定如何处理闰年。这是我的方法,我认为它非常易读,并且能够在不使用任何“特殊情况”逻辑的情况下大踏步前进。

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

威辛 http://github.com/radar/dotiw

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

生产

Jack is 60 years old

你可以用红宝石 adroit-age

它也适用于闰年。

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

更新

require 'adroit-age'

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

age = dob.find_age
#=> 23

基于与@fmc相似的推理,我得出了以下结论 首先,计算今天一年和生日一年的差额。然后,将其和到生日并检查结果日期:如果大于今天,则将diff减少1。 在Rails应用程序中使用,因为它依赖 ActiveSupport 的 years 方法

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

与fm的想法相同,但有一个简化的if语句。显然,您可以添加第二个参数,而不是使用当前时间。

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

我认为这总是有效的,即使对于一个在闰日附近过生日的人:

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'))

在上面的示例调用中,使用此方法计算的年龄是42,这是正确的,尽管1968年是闰年,生日接近闰日。

另外,这样就不需要创建局部变量。

这个怎么样?

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

比如:

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

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