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。

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。

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

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)`

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 }}

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。