如何在不同的语句中循环多个条件，如果条件不满足则使其循环？

Question

我需要专门针对我的代码的帮助。 output 似乎不是我想要的。 如果我尝试 while 循环它，并且值是错误的，它仍然会继续。 这里有很多错误，我不知道如何修复。

weapon_2 = "axe"
weapon_3 = "spiked club"

weapon_1 = weapon_1.strip().lower()
weapon_2 = weapon_1.strip().lower()
weapon_3 = weapon_1.strip().lower()

weapon = input ("Please select a weapon: Sword, Axe, Spiked Club.\n")
weapon = weapon.strip().lower()



if weapon == weapon_1:
    print("Nice choice! I would pick {} too!".format(weapon))
    
elif weapon == weapon_2:
    print("Nice choice! I would pick {} too!".format(weapon))
    
elif weapon == weapon_3:
    print("Nice choice! I would pick {} too!".format(weapon))

while weapon != (weapon_1) or (weapon_2) or (weapon_3):
    print("Invalid option.")
    weapon = input ("Please select a weapon: Sword, Axe, Spiked Club.\n")    

    

Answer 1

代替：

while weapon:= (weapon_1) or (weapon_2) or (weapon_3):

和：

while weapon not in (weapon_1, weapon_2, weapon_3):

or正在将(weapon_2)评估为 boolean，它看到一个str因此评估为True并且while循环继续运行。

您还应该使打印语句中的选项在文本上与选项相同，否则用户将键入“Axe”，它与“axe”不匹配并被告知这是一个无效选项。

Answer 2

需要对 while 循环进行一些改进以保持“循环”

weapon_1, weapon_2, weapon_3  = "sword", "axe", "spiked club"

while True:
    weapon = input("Please select a weapon: Sword, Axe, Spiked Club.\n").strip().lower()
    
    if weapon == weapon_1:
        print(f"Nice choice! I would pick {weapon_1} too!")
    elif weapon == weapon_2:
        print(f"Nice choice! I would pick {weapon_2} too!")
    elif weapon == weapon_3:
        print(f"Nice choice! I would pick {weapon_3} too!")
    elif weapon == 'stop':
        break
    else:
        print("Invalid Input")

Answer 3

weapons = ["sword", "axe", "spiked club"]

while True:

    # input
    user_input = input("Please select a weapon: Sword, Axe, Spiked Club. ('stop' to exit)\n")
    user_input_lower = user_input.strip().lower()
    if user_input_lower == 'stop':
        break

    # weapon precessing
    weapon = user_input_lower
    if weapon in weapons:
        print(f"Nice choice! I would pick {user_input} too!")
    else:
        print("Invalid Input")