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}')

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

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)

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。

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

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)