如何在python中打印错误？

我最近刚刚完成了一个非常基本的故障排除系统的一些代码。我只是想添加一些验证，这样，如果用户在任何问题上输入了不是“是”或“否”的内容，那么它将在shell中打印“无效输入”，但不确定如何准确地执行。有人能帮我吗？非常感谢

我的代码：

prompt = "> "

print ("screen question one here")
screen = input(prompt)
if screen == "yes":
    print ("screen question two here")
    screen2 = input(prompt)
    if screen2 == "yes":
        print ("screen question three here")
        screen3 = input(prompt)
        if screen3 == "yes":
            print ("screen advice one here")
        elif screen3 == "no":
            print ("screen adivce two here")
    elif screen2 == "no":
        print ("camera question one here")
        camera = input(prompt)
        if camera == "yes":
            print ("camera question two here")
            camera2 = input(prompt)
            if camera2 == "yes":
                print ("camera advice one here")
            elif camera2 == "no":
                print ("camera advice two here")
        elif camera == "no":
            print ("water question one here")
            water = input(prompt)
            if water == "yes":
                print ("water question two here")
                water2 = input(prompt)
                if water2 == "yes":
                    print ("water advice one here")
                elif water2 == "no":
                    print ("water advice two here")
            elif water == "no":
                print ("buttons question one here")
                buttons = input(prompt)
                if buttons == "yes":
                    print ("buttons advice one here")
                elif buttons == "no":
                    print ("buttons advice two here")  
elif screen == "no":
    print ("battery question one here")
    battery = input(prompt)
    if battery == "yes":
        print ("battery question two here")
        battery2 = input(prompt)
        if battery2 == "yes":
            print ("battery advice one here")
        elif battery2 == "no":
            print ("battery advice two here")
    elif battery == "no":
        print ("wifi question one here")
        wifi = input(prompt)
        if wifi == "yes":
            print ("wifi advice one here")
        elif wifi == "no":
            print ("wifi advice two here")

---

这里有一种方法，你可以这样做

定义一个函数，从用户处获取是或否。人们倾向于重复这个问题，直到用户给你一个合适的回答：这就是这个函数的作用

def yesorno(question):
    while True:
        print(question)
        answer = input('> ').strip().lower()
        if answer in ('y', 'yes'):
            return True
        if answer in ('n', 'no'):
            return False
        print("Invalid input")

另一方面，如果您只想在获得无效输入时退出脚本，可以执行以下操作：

def yesorno(question):
    print(question)
    answer = input('> ').strip().lower()
    if answer in ('y', 'yes'):
        return True
    if answer in ('n', 'no'):
        return False
    exit('Invalid input')

exit

将直接退出整个程序。这不一定是我推荐的

无论哪种方式，您都可以这样使用

yesorno

：

if yesorno("Question 1"):
    # User answered yes
    if yesorno("Question 2A"):
        # User answered yes
    else:
        # User answered no
else:
    # User answered no
    if yesorno("Question 2B"):
        ...

---

首先，编写一个函数，从用户那里得到是或否。集中精力让它按你想要的方式工作，然后你就可以用它来回答所有不同的问题了！但我想你早就知道了。还在挣扎，它要么解决了一些问题，要么根本没有。你能帮忙吗@克尔伍德：这工作非常好！请问.strip（）在这段代码中的作用是什么？多谢各位@khelwood从字符串中删除周围的空格，因此如果用户键入

“yes”

，它将转换为

“yes”

。