python - 如何检查用户的输入是否有效 Python

Question

这是我的练习:获取用户的名字、姓氏和出生年份。根据姓名(第一个字母)创建首字母并计算用户的年龄。

我写了一个脚本:

def age_name():
    # var= firstName ,lastName ,year
    is_running=True
    firstName =input("What is your name?").lower()
    while is_running==True:
        if firstName.isalpha() and len(firstName)>0:
            lastName = input("What is your last name?").lower()
            if lastName.isalpha() and len(lastName) > 0:
                year = input("What is your birth year?")
                if year.isnumeric() or len(year)==4:
                    year=int(year)
                    print("Your initials are {0}{1} and your age is {2}.".format(firstName[0],lastName[0],2018-year))
                    #print("Your initials are ",firstName[0],lastName[0],"and your age is ",str(2018-year))
                else:
                    print("invalid year. please type your birth year")
                    year = input("What is your birth year?")
            else:
                print("invalid last name. please type your last name")
                lastName = input("What is you last name?").lower()
        else:
            print("invalid name. please type your name")
            firstName = input("What is you name?").lower()

age_name()

我已经测试了代码，这是我得到的:

What is your name?p5
invalid name. please type your name
What is you name?peter
What is your last name?p5
invalid last name. please type your last name
What is you last name?pen
What is your last name?pen
What is your birth year?p5
invalid year. please type your birth year
What is your birth year?10
What is your last name?pen
What is your birth year?1800
Your initials are pp and your age is 218.
What is your last name?

首先，我认为我的代码有点重复——如果有人想缩短它的话。第二个也是最大的问题 - 我不断收到姓氏问题。我的错误在哪里？

Answer 1

我认为最好的可读性同时最不容易出错的方法是如下所示。输入线每条仅编程一次，一切都取决于 response_invalid 的状态。

def age_name():
    response_invalid = True
    while response_invalid:
        firstName = input("What is your name?").lower()
        if firstName.isalpha() and len(firstName)>0:
            response_invalid = False
        else:
            print("invalid name. please type your name")
    response_invalid = True
    while response_invalid:
        lastName = input("What is your last name?").lower()
        if lastName.isalpha() and len(lastName) > 0:
            response_invalid = False
        else:
            print("invalid last name. please type your last name")
    response_invalid = True
    while response_invalid:
        year = input("What is your birth year?")
        if year.isnumeric() or len(year)==4:
            response_invalid = False
        else:
            print("invalid year. please type your birth year")
    year=int(year)
    print("Your initials are {0}{1} and your age is {2}.".format(firstName[0],lastName[0],2018-year))

然而，这是一种非常常见的实现方式的冗长变体，我认为这是最短的，但有点粗鲁......:

def age_name():
    while True:
        firstName = input("What is your name?").lower()
        if firstName.isalpha() and len(firstName)>0:
            break
        print("invalid name. please type your name")
    while True:
        lastName = input("What is your last name?").lower()
        if lastName.isalpha() and len(lastName) > 0:
            break
        print("invalid last name. please type your last name")
    while True:
        year = input("What is your birth year?")
        if year.isnumeric() or len(year)==4:
            break
        print("invalid year. please type your birth year")
    year=int(year)
    print("Your initials are {0}{1} and your age is {2}.".format(firstName[0],lastName[0],2018-year))