如果用户未选择1或2,我希望代码说“请输入1或2继续”,则该部分有效。但是,如果用户输入“ 6”,则会按要求输入“请输入1或2以继续”,但是如果在无效输入后直接输入了有效输入,则代码将无法正确显示。
我试图在没有需求函数的情况下执行此操作,但是似乎并没有任何效果。
def requirement():
choice = ""
while choice != "1" and choice != "2":
choice = input ("Please enter 1 or 2 to continue.\n")
if choice == "1" and choice == "2":
return choice
def intro():
print ("Enter 1 to enter the cave\n")
print ("Enter 2 to explore the river\n")
play_again = input ("What would you like to do?\n")
if play_again in "1":
print ("You win!")
elif play_again in "2":
print ("YOU LOSE")
print ("Thanks for playing!")
exit()
else:
requirement()
intro()
答案 0 :(得分:0)
def intro():
print ("Enter 1 to enter the cave\n")
print ("Enter 2 to explore the river\n")
play_again = input ("What would you like to do?\n")
return play_again
def game(choice):
if choice == "1":
print ("You win!")
elif choice == "2":
print ("YOU LOSE")
print ("Thanks for playing!")
exit()
else:
choice = input ("Please enter 1 or 2 to continue.\n")
game(choice)
game(intro())
else
语句已经处理了是否输入1或2,因此不需要requirement
函数。