Python 或字符串操作

Question

I am working on an entry-level project of an automated coffee shop and I ran into a small problem with my code. I have this code if order != "mocha" or "frappuccino" or "cappuccino": in order to filter out selections at the very first choice. But when I try to run my code, the response from the function (yes_or_no_question) also seem to go through the if statement mentioned above and I can not figure out why. I do suspect that it is a problem with the OR operation and strings if not a logical error with the way I have my code set up. I had included my test run errors below, thank you all for your valuable time to help out beginners.

#side functions
def ask_amount_m():
    amount = int(input("How many would you like?"))
    response_1 = "{} mocha coming up".format(amount)
    print(response_1)
    return amount
def ask_amount_f():
    amount = int(input("How many would you like?"))
    response_2 = "{} frappuccino coming up".format(amount)
    print(response_2)
    return amount
def ask_amount_c():
    amount = int(input("How many would you like?"))
    response_3 = "{} cappuccino coming up".format(amount)
    print(response_3)
    return amount
**def yes_or_no_question():**
    x = input("Would you like to order anything else?")
    if x == "yes":
        first_step()
    if x == "no":
        response_bye = "Have a nice day!"
        print(response_bye)

#main code starts here
def first_step():
    order = input("What would you like to order today?")
    if order == "mocha":
        ask_amount_m()
        yes_or_no_question()

    if order == "frappuccino":
        ask_amount_f()
        yes_or_no_question()

    if order == "cappuccino":
        ask_amount_c()
        yes_or_no_question()

    **if order != "mocha" or "frappuccino" or "cappuccino":**
        print("Try again")
        first_step()

first_step()

Test run error example:
What would you like to order today?kk
Try again
What would you like to order today?kjgg;]
Try again
What would you like to order today?frappuccino
How many would you like?234
234 frappuccino coming up
Would you like to order anything else?no
Have a nice day!
Try again
What would you like to order today?

Answer 1

The or doesn't work like this. Replace the line with the following:

if order not in ("mocha", "frappuchino", "cappuchino"):

Answer 2

试试这个： if order != "mocha" or order != "frappuccino" or order != "cappuccino":**

Answer 3

You could use the in operator:

if order not in ["mocha", "frappuccino", "cappuccino"]:
    # better grab some tea

Answer 4

Try not to use recursion in python. Consider something like like this code.

def ask_amount(kind_of_coffee):
    amount = int(input("How many would you like?"))
    print("{} {} coming up".format(amount, kind_of_coffee))
    return amount

def is_finished():
    x = input("Would you like to order anything else?")
    if x == "no":
        response_bye = "Have a nice day!"
        print(response_bye)
        return True
    return False

ACCEPTED_BEVRAGES = ["mocha", "frappuccino", "cappuccino"]

#main code starts here
def first_step():
    finished = False

    while not finished:
        order = input("What would you like to order today?")
        if order in ACCEPTED_BEVRAGES:
            # The order is for a "mocha", a "frappuccino", or a "cappuccino"
            ask_amount(order)

            finished = is_finished()
        else:
            print("Try again")


first_step()

How ever, you had a big problem in your code that is worth pointing out

if order != "mocha" or "frappuccino" or "cappuccino" will not work in python. You need to be specific and write it like if order != "mocha" or order != "frappuccino" or order != "cappuccino" . Even better would be to check for absence in list, see updated code for this.

Also, you really don't have to use so many different functions that does exactly the same thing. Fuctions can accept parameters too.