基于用户输入的基本 Python Function 选择系统

Question

这是我一直在做的一个基本项目，一个基于文本的冒险，（我知道是原创的）但我想向它添加一个库存 select 系统。

我想给出 4 个可能列表之一的 output，从用户输入中，他们可以询问每个列表和 select 它。 他们可以根据需要取消，function 将循环到开头。

它应该返回一个列表和一个数字的 output，但它似乎没有 output 除了＃1。 任何人都可以看到有什么问题吗？

另外，在任何人说之前，我知道它的垃圾代码，但这是我的第一个项目，任何简化和压缩它的建议将不胜感激！

inv1 = ["Sword", "Flask of Water", "Pebble", "Sheild"]
inv2 = ["Bow", "Quivver of Arrows", "Gold Necklace"]
inv3 = ["Dagger", "Bottle of Poison", "Throwing Knives"]
inv4 = ["Spellbook: Cast Fireball", "Spellbook: Heal", "Spellbook: Control Skelleton"]
emptyinvt = []
z = 0

def invt(a):
    if a == "1":
        print (inv1)
        ans1 = input("Do you want to take loadout 1? ")
        if ans1 == "yes":
            return 1
        elif ans1 == "no":
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
        else:
            print ("Not a option!")
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
    elif a == "2":
        print (inv2)
        ans2 = input("Do you want to take loadout 2? ")
        if ans2 == "yes":
            return 2
        elif ans2 == "no":
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
        else:
            print ("Not a option!")
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))

    elif a == "3":
        print (inv3)
        ans3 = input("Do you want to take loadout 3? ")
        if ans3 == "yes":
            return 3
        elif ans3 == "no":
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
        else:
            print ("Not a option!")
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))

    elif a == "4":
        print (inv4)
        ans4 = input("Do you want to take loadout 4? ")
        if ans4 == "yes":
            return 4
        elif ans4 == "no":
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
        else:
            print ("Not a option!")
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
    else:
            print ("Not a option!")
            invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))

int_invent = invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))
print (int_invent)

print ("player INVT " + str(int_invent))

if int_invent == 1:
    plrinvt = list(set(inv1 + emptyinvt))
elif int_invent == 2:
    plrinvt = list(set(inv2 + emptyinvt))
elif int_invent == 3:
    plrinvt = list(set(inv3 + emptyinvt))
elif int_invent == 4:
    plrinvt = list(set(inv4 + emptyinvt))

print ("Player Inventory %d has been selected" % int_invent)
print ("It contains: " + str(plrinvt))

Answer 1

如果第一个输入有效，您的代码工作正常。 但是，您的递归调用失败，因为返回值为 None。 当用户在输入后说不并且给出无效值时，会发生递归调用。 您递归调用 function 但如果选择主线程，它不会返回选择的值。 一种解决方案是在主线程上使用条件循环来确保选择正确的序列。 你可以循环你的 function 如下：

inv1 = ["Sword", "Flask of Water", "Pebble", "Sheild"]
inv2 = ["Bow", "Quivver of Arrows", "Gold Necklace"]
inv3 = ["Dagger", "Bottle of Poison", "Throwing Knives"]
inv4 = ["Spellbook: Cast Fireball", "Spellbook: Heal", "Spellbook: Control Skelleton"]
emptyinvt = []
z = 0

def invt(a):
    if a == "1":
        print (inv1)
        ans1 = input("Do you want to take loadout 1? ")
        if ans1 == "yes":
            return 1
        elif ans1 == "no":
            return None
        else:
            print ("Not a option!")
            return None
    elif a == "2":
        print (inv2)
        ans2 = input("Do you want to take loadout 2? ")
        if ans2 == "yes":
            return 2
        elif ans2 == "no":
            return None
        else:
            print ("Not a option!")
            return None

    elif a == "3":
        print (inv3)
        ans3 = input("Do you want to take loadout 3? ")
        if ans3 == "yes":
            return 3
        elif ans3 == "no":
            return None
        else:
            print ("Not a option!")
            return None

    elif a == "4":
        print (inv4)
        ans4 = input("Do you want to take loadout 4? ")
        if ans4 == "yes":
            return 4
        elif ans4 == "no":
           return None #we return None so that the while loop works
        else:
            print ("Not a option!")
            return None
    else:
            print ("Not a option!")
            return None

int_invent = None
while(int_invent is None): #If there was a problem in invt it just relaunches it with the same query
    int_invent = invt(input("Your king offers you three bundles of tools for your journey, which do you take?"))


print ("player INVT " + str(int_invent))

if int_invent == 1:
    plrinvt = list(set(inv1 + emptyinvt))
elif int_invent == 2:
    plrinvt = list(set(inv2 + emptyinvt))
elif int_invent == 3:
    plrinvt = list(set(inv3 + emptyinvt))
elif int_invent == 4:
    plrinvt = list(set(inv4 + emptyinvt))
print(int_invent)
print ("Player Inventory %d has been selected" % int_invent)
print ("It contains: " + str(plrinvt))

Answer 2

我有一个建议，我已经尝试过你的游戏。 你应该让它对用户更友好。 例如

'Your king offers you three bundles of tools for your journey, which do you take?'

，您应该在最后添加可能的答案。 这样玩起来会容易一些。 例子：

'Your king offers you three bundles of tools for your journey, which do you take? (1,2,3)'. 

我试图帮助你，但我不明白你的问题和你想要做什么。 请详细说明。