不幸的是，您不能只使用时间数据，因为它使用的最大单位是日，闰年将使您的计算无效。因此，让我们找到年数，然后如果最后一年没有满，就按1调整:

from datetime import date
birth_date = date(1980, 5, 26)
years = date.today().year - birth_date.year
if (datetime.now() - birth_date.replace(year=datetime.now().year)).days >= 0:
    age = years
else:
    age = years - 1

Upd:

这个解决方案在2月29日开始时确实会导致一个异常。以下是正确的检查:

from datetime import date
birth_date = date(1980, 5, 26)
today = date.today()
years = today.year - birth_date.year
if all((x >= y) for x,y in zip(today.timetuple(), birth_date.timetuple()):
   age = years
else:
   age = years - 1

Upd2:

将多次调用now()称为性能损失是荒谬的，除非在极端特殊的情况下，否则这无关紧要。使用变量的真正原因是数据不一致的风险。

在这种情况下，最经典的问题是如何对待2月29日出生的人。例如:你必须年满18岁才能投票、开车、买酒等等。如果你出生在2004-02-29，你被允许做这些事情的第一天是哪一天:2022-02-28，或2022-03-01?AFAICT，大多数是前者，但一些扫兴的人可能会说是后者。

下面的代码迎合了那一天出生的0.068%(大约)人口:

def age_in_years(from_date, to_date, leap_day_anniversary_Feb28=True):
    age = to_date.year - from_date.year
    try:
        anniversary = from_date.replace(year=to_date.year)
    except ValueError:
        assert from_date.day == 29 and from_date.month == 2
        if leap_day_anniversary_Feb28:
            anniversary = datetime.date(to_date.year, 2, 28)
        else:
            anniversary = datetime.date(to_date.year, 3, 1)
    if to_date < anniversary:
        age -= 1
    return age

if __name__ == "__main__":
    import datetime

    tests = """

    2004  2 28 2010  2 27  5 1
    2004  2 28 2010  2 28  6 1
    2004  2 28 2010  3  1  6 1

    2004  2 29 2010  2 27  5 1
    2004  2 29 2010  2 28  6 1
    2004  2 29 2010  3  1  6 1

    2004  2 29 2012  2 27  7 1
    2004  2 29 2012  2 28  7 1
    2004  2 29 2012  2 29  8 1
    2004  2 29 2012  3  1  8 1

    2004  2 28 2010  2 27  5 0
    2004  2 28 2010  2 28  6 0
    2004  2 28 2010  3  1  6 0

    2004  2 29 2010  2 27  5 0
    2004  2 29 2010  2 28  5 0
    2004  2 29 2010  3  1  6 0

    2004  2 29 2012  2 27  7 0
    2004  2 29 2012  2 28  7 0
    2004  2 29 2012  2 29  8 0
    2004  2 29 2012  3  1  8 0

    """

    for line in tests.splitlines():
        nums = [int(x) for x in line.split()]
        if not nums:
            print
            continue
        datea = datetime.date(*nums[0:3])
        dateb = datetime.date(*nums[3:6])
        expected, anniv = nums[6:8]
        age = age_in_years(datea, dateb, anniv)
        print datea, dateb, anniv, age, expected, age == expected

输出如下:

2004-02-28 2010-02-27 1 5 5 True
2004-02-28 2010-02-28 1 6 6 True
2004-02-28 2010-03-01 1 6 6 True

2004-02-29 2010-02-27 1 5 5 True
2004-02-29 2010-02-28 1 6 6 True
2004-02-29 2010-03-01 1 6 6 True

2004-02-29 2012-02-27 1 7 7 True
2004-02-29 2012-02-28 1 7 7 True
2004-02-29 2012-02-29 1 8 8 True
2004-02-29 2012-03-01 1 8 8 True

2004-02-28 2010-02-27 0 5 5 True
2004-02-28 2010-02-28 0 6 6 True
2004-02-28 2010-03-01 0 6 6 True

2004-02-29 2010-02-27 0 5 5 True
2004-02-29 2010-02-28 0 5 5 True
2004-02-29 2010-03-01 0 6 6 True

2004-02-29 2012-02-27 0 7 7 True
2004-02-29 2012-02-28 0 7 7 True
2004-02-29 2012-02-29 0 8 8 True
2004-02-29 2012-03-01 0 8 8 True

from datetime import date

def calculate_age(born):
    today = date.today()
    try: 
        birthday = born.replace(year=today.year)
    except ValueError: # raised when birth date is February 29 and the current year is not a leap year
        birthday = born.replace(year=today.year, month=born.month+1, day=1)
    if birthday > today:
        return today.year - born.year - 1
    else:
        return today.year - born.year

更新:使用丹尼的解决方案，效果更好

由于我没有看到正确的实现，我以这种方式重新编码了我的…

    def age_in_years(from_date, to_date=datetime.date.today()):
  if (DEBUG):
    logger.debug("def age_in_years(from_date='%s', to_date='%s')" % (from_date, to_date))

  if (from_date>to_date): # swap when the lower bound is not the lower bound
    logger.debug('Swapping dates ...')
    tmp = from_date
    from_date = to_date
    to_date = tmp

  age_delta = to_date.year - from_date.year
  month_delta = to_date.month - from_date.month
  day_delta = to_date.day - from_date.day

  if (DEBUG):
    logger.debug("Delta's are : %i  / %i / %i " % (age_delta, month_delta, day_delta))

  if (month_delta>0  or (month_delta==0 and day_delta>=0)): 
    return age_delta 

  return (age_delta-1)

如果你出生在2月29日，你就认为自己在2月28日是“18岁”，这是错误的。 交换边界可以省略…这只是我的代码的个人方便:)

from datetime import date

days_in_year = 365.2425    
age = int((date.today() - birth_date).days / days_in_year)

在Python 3中，你可以对datetime.timedelta执行除法:

from datetime import date, timedelta

age = (date.today() - birth_date) // timedelta(days=365.2425)

进口日期时间

def age(date_of_birth):
    if date_of_birth > datetime.date.today().replace(year = date_of_birth.year):
        return datetime.date.today().year - date_of_birth.year - 1
    else:
        return datetime.date.today().year - date_of_birth.year

在你的情况下:

import datetime

# your model
def age(self):
    if self.birthdate > datetime.date.today().replace(year = self.birthdate.year):
        return datetime.date.today().year - self.birthdate.year - 1
    else:
        return datetime.date.today().year - self.birthdate.year

from datetime import date

def age(birth_date):
    today = date.today()
    y = today.year - birth_date.year
    if today.month < birth_date.month or today.month == birth_date.month and today.day < birth_date.day:
        y -= 1
    return y

考虑到int(True)是1,int(False)是0，元组比较从左到右，这可以做得更简单:

from datetime import date

def calculate_age(born):
    today = date.today()
    return today.year - born.year - ((today.month, today.day) < (born.month, born.day))

最简单的方法是使用python-dateutil

import datetime

import dateutil

def birthday(date):
    # Get the current date
    now = datetime.datetime.utcnow()
    now = now.date()

    # Get the difference between the current date and the birthday
    age = dateutil.relativedelta.relativedelta(now, date)
    age = age.years

    return age

下面是一个用年、月或日来计算一个人年龄的方法。

假设一个人的出生日期是2012-01-17T00:00:00 因此，他的年龄在2013-01-16T00:00:00将是11个月

或者他出生在2012-12-17T00:00:00， 他的年龄在2013-01-12T00:00:00将是26天

或者他出生在2000-02-29T00:00:00， 他的年龄在2012-02-29T00:00:00将是12岁

您将需要导入datetime。

代码如下:

def get_person_age(date_birth, date_today):

"""
At top level there are three possibilities : Age can be in days or months or years.
For age to be in years there are two cases: Year difference is one or Year difference is more than 1
For age to be in months there are two cases: Year difference is 0 or 1
For age to be in days there are 4 possibilities: Year difference is 1(20-dec-2012 - 2-jan-2013),
                                                 Year difference is 0, Months difference is 0 or 1
"""
years_diff = date_today.year - date_birth.year
months_diff = date_today.month - date_birth.month
days_diff = date_today.day - date_birth.day
age_in_days = (date_today - date_birth).days

age = years_diff
age_string = str(age) + " years"

# age can be in months or days.
if years_diff == 0:
    if months_diff == 0:
        age = age_in_days
        age_string = str(age) + " days"
    elif months_diff == 1:
        if days_diff < 0:
            age = age_in_days
            age_string = str(age) + " days"
        else:
            age = months_diff
            age_string = str(age) + " months"
    else:
        if days_diff < 0:
            age = months_diff - 1
        else:
            age = months_diff
        age_string = str(age) + " months"
# age can be in years, months or days.
elif years_diff == 1:
    if months_diff < 0:
        age = months_diff + 12
        age_string = str(age) + " months" 
        if age == 1:
            if days_diff < 0:
                age = age_in_days
                age_string = str(age) + " days" 
        elif days_diff < 0:
            age = age-1
            age_string = str(age) + " months"
    elif months_diff == 0:
        if days_diff < 0:
            age = 11
            age_string = str(age) + " months"
        else:
            age = 1
            age_string = str(age) + " years"
    else:
        age = 1
        age_string = str(age) + " years"
# The age is guaranteed to be in years.
else:
    if months_diff < 0:
        age = years_diff - 1
    elif months_diff == 0:
        if days_diff < 0:
            age = years_diff - 1
        else:
            age = years_diff
    else:
        age = years_diff
    age_string = str(age) + " years"

if age == 1:
    age_string = age_string.replace("years", "year").replace("months", "month").replace("days", "day")

return age_string

以上代码中使用的一些额外函数是:

def get_todays_date():
    """
    This function returns todays date in proper date object format
    """
    return datetime.now()

And

def get_date_format(str_date):
"""
This function converts string into date type object
"""
str_date = str_date.split("T")[0]
return datetime.strptime(str_date, "%Y-%m-%d")

现在，我们必须为get_date_format()提供像2000-02-29T00:00:00这样的字符串

它将把它转换为日期类型对象，该对象将被提供给get_person_age(date_birth, date_today)。

函数get_person_age(date_birth, date_today)将返回字符串格式的年龄。

如果你想用django模板打印在页面中，那么下面的代码就足够了:

{{ birth_date|timesince }}

扩展了Danny的解决方案，但有各种各样的方法来报告年轻人的年龄(注意，今天是datetime.date(2015,7,17)):

def calculate_age(born):
    '''
        Converts a date of birth (dob) datetime object to years, always rounding down.
        When the age is 80 years or more, just report that the age is 80 years or more.
        When the age is less than 12 years, rounds down to the nearest half year.
        When the age is less than 2 years, reports age in months, rounded down.
        When the age is less than 6 months, reports the age in weeks, rounded down.
        When the age is less than 2 weeks, reports the age in days.
    '''
    today = datetime.date.today()
    age_in_years = today.year - born.year - ((today.month, today.day) < (born.month, born.day))
    months = (today.month - born.month - (today.day < born.day)) %12
    age = today - born
    age_in_days = age.days
    if age_in_years >= 80:
        return 80, 'years or older'
    if age_in_years >= 12:
        return age_in_years, 'years'
    elif age_in_years >= 2:
        half = 'and a half ' if months > 6 else ''
        return age_in_years, '%syears'%half
    elif months >= 6:
        return months, 'months'
    elif age_in_days >= 14:
        return age_in_days/7, 'weeks'
    else:
        return age_in_days, 'days'

示例代码:

print '%d %s' %calculate_age(datetime.date(1933,6,12)) # >=80 years
print '%d %s' %calculate_age(datetime.date(1963,6,12)) # >=12 years
print '%d %s' %calculate_age(datetime.date(2010,6,19)) # >=2 years
print '%d %s' %calculate_age(datetime.date(2010,11,19)) # >=2 years with half
print '%d %s' %calculate_age(datetime.date(2014,11,19)) # >=6 months
print '%d %s' %calculate_age(datetime.date(2015,6,4)) # >=2 weeks
print '%d %s' %calculate_age(datetime.date(2015,7,11)) # days old

80 years or older
52 years
5 years
4 and a half years
7 months
6 weeks
7 days

正如@[Tomasz Zielinski]和@ williams所建议的那样，python-dateutil可以只做5行。

from dateutil.relativedelta import *
from datetime import date
today = date.today()
dob = date(1982, 7, 5)
age = relativedelta(today, dob)

>>relativedelta(years=+33, months=+11, days=+16)`

为了便于阅读和理解，稍微修改了Danny的解决方案

    from datetime import date

    def calculate_age(birth_date):
        today = date.today()
        age = today.year - birth_date.year
        full_year_passed = (today.month, today.day) < (birth_date.month, birth_date.day)
        if not full_year_passed:
            age -= 1
        return age

延伸到丹尼·w·阿代尔回答，得到月也

def calculate_age(b):
    t = date.today()
    c = ((t.month, t.day) < (b.month, b.day))
    c2 = (t.day< b.day)
    return t.year - b.year - c,c*12+t.month-b.month-c2

import datetime

今天的日期

td=datetime.datetime.now().date() 

你的出生年月日

bd=datetime.date(1989,3,15)

你的年龄

age_years=int((td-bd).days /365.25)

你可以使用Python 3来完成这一切。只需运行下面的代码就可以了。

# Creating a variables:

greeting = "Hello, "
name = input("what is your name?")
birth_year = input("Which year you were born?")
response = "Your age is "

# Converting string variable to int:

calculation = 2020 - int(birth_year) 


# Printing:

print(f'{greeting}{name}. {response}{calculation}')

serializers.py

age = serializers.SerializerMethodField('get_age')

class Meta:
    model = YourModel
    fields = [..,'','birthdate','age',..]


import datetime

def get_age(self, instance):
    return datetime.datetime.now().year - instance.birthdate.year

一个比@DannyWAdairs稍微优雅一点的解决方案可能是使用.timetuple()方法[Python-doc]:

from datetime import date

def calculate_age(born):
    today = date.today()
    return today.year - born.year - (today.timetuple()[1:3] < born.timetuple()[1:3])

你可以很容易地使用这个来进一步推广它，将其粒度增加到秒，这样，如果它大于或等于当天的秒数，年龄就会增加，例如born是一个datetime对象:

from datetime import datetime

def calculate_age_with_seconds(born):
    today = datetime.now()
    return today.year - born.year - (today.timetuple()[1:6] < born.timetuple()[1:6])

这对于date或datetime对象都适用。