如果他们说不，我如何提示用户退出并重新启动程序[重复]

Question

有人可以帮忙说明如何提示用户退出，如果他们拒绝，则返回并重新运行程序？我对此感到非常头疼，而互联网也没有帮助。

这是我的代码：

print("Hello, this program will ask for two numbers, then show the product 
of those numbers that many times")

value = input("Please enter number")

value = int(value)

value2 = input("Please enter a second number")

value2 = int(value2)

for i in range(value*value2):
    print(value*value2)

    while True:
        answer = input("Do you wih to exit? Enter Yes or No:")
        if answer in ('No', 'Yes'):
            break
        print ("Invalid input.")
    if answer == 'No':
        continue
    else:
        print ("Goodbye")
        break

它只是乘法，然后显示和本身一样多次，但我不知道如何提示用户退出，如果他们说不，则再次运行程序。我听说过将整个代码放在一个循环中，并且我尝试过，但无济于事。我能得到一些帮助吗？

Answer 1

根据所写的答案，但未编写函数： 这里的关键是您需要循环直到用户输入“是”。你的while 循环永远不会是假的，所以你会永远提示。这是另一个解决方案（尽管@Ricky Kim 有一个更 Pythonic 的答案）

answer=""
while answer!= "yes":
    print("Hello, this program will ask for two numbers, then show the product \
    of those numbers that many times")

    value = input("Please enter number")

    value = int(value)

    value2 = input("Please enter a second number")

    value2 = int(value2)

    for i in range(value*value2):
        print(value*value2)
    #use lower so that the user can enter "Yes" or "yes"
    answer = str.lower(input("Do you wih to exit? Enter Yes or No:"))
print ("Goodbye")

Answer 2

您可以将其设为递归函数，并在用户回答否时执行循环。

例如：

def ask():
    print("Hello, this program will ask for two numbers, then show the product of those numbers that many times")

    value = input("Please enter number")

    value = int(value)

    value2 = input("Please enter a second number")

    value2 = int(value2)

    for i in range(value * value2):
        print(value * value2)
    answer = input("Do you wih to exit? Enter Yes or No:")
    if answer not in ('No', 'Yes'):
        print("Invalid input.")
    if answer == 'No':
        ask()
    else:
        print("Goodbye")
ask()