12 回答

• 5

  您是否想要年龄，因为人们通常会理解它，或者您是否正在寻找精确的时间测量？如果是前者，就没有必要担心闰年等并发症 . 你只需要计算几年的差异，如果这个人今年没有过生日就减少它 . 如果是后者，您可以将经过的秒数转换为年，如其他答案所示 .

  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)
  

  回复于 2024-05-06T03:33:34+08:00
• 0

  与http://github.com/radar/dotiw

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

  生产环境

  Jack is 60 years old
  

  回复于 2024-05-06T03:33:34+08:00
• 5

  我认为这将永远有效，即使是在闰日附近有生日的人：

  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年是闰年而生日是在闰日附近 .

  另外，这种方式不需要创建局部变量 .

  回复于 2024-05-06T03:33:34+08:00
• 2

  基于与@FMc First类似的推理，我提出了以下内容，计算今天's year and birthday'年之间的差异 . 然后，将它加到生日并检查结果日期：如果它大于今天，将差异减少1.在Rails应用程序中使用，因为它依赖于 ActiveSupport 的 years 方法

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

  回复于 2024-05-06T03:33:34+08:00
• 2

  你可以使用红宝石宝石adroit-age

  它适用于闰年也..

  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
  

  回复于 2024-05-06T03:33:34+08:00
• 0

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

  回复于 2024-05-06T03:33:34+08:00
• 1

  怎么样的：

  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
  

  回复于 2024-05-06T03:33:34+08:00
• 1

  处理闰年的方法

  无论何时计算自某个日期起的经过年限，您都必须决定如何处理闰年 . 这是我的方法，我认为这是非常易读的，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
  

  回复于 2024-05-06T03:33:34+08:00
• 4

  与FM相同的想法，但使用简化的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
  

  回复于 2024-05-06T03:33:34+08:00
• 1

  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
  

  回复于 2024-05-06T03:33:34+08:00
• 39

  我有一个名为dotiw的gem /插件，它有一个 distance_of_time_in_words_hash ，它将返回一个像： { :years => 59, :months => 11, :days => 27 } 这样的哈希 . 如果它接近某个限制，你可以解决这个问题 .

  回复于 2024-05-06T03:33:34+08:00
• 1

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

  回复于 2024-05-06T03:33:34+08:00