如何在不同的語句中循環多個條件，如果條件不滿足則使其循環？

Question

我需要專門針對我的代碼的幫助。 output 似乎不是我想要的。 如果我嘗試 while 循環它，並且值是錯誤的，它仍然會繼續。 這里有很多錯誤，我不知道如何修復。

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正在將(weapon_2)評估為 boolean，它看到一個str因此評估為True並且while循環繼續運行。

您還應該使打印語句中的選項在文本上與選項相同，否則用戶將鍵入“Axe”，它與“axe”不匹配並被告知這是一個無效選項。

Answer 2

需要對 while 循環進行一些改進以保持“循環”

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")