基於用戶輸入的基本 Python Function 選擇系統

Question

這是我一直在做的一個基本項目，一個基於文本的冒險，（我知道是原創的）但我想向它添加一個庫存 select 系統。

我想給出 4 個可能列表之一的 output，從用戶輸入中，他們可以詢問每個列表和 select 它。 他們可以根據需要取消，function 將循環到開頭。

它應該返回一個列表和一個數字的 output，但它似乎沒有 output 除了＃1。 任何人都可以看到有什么問題嗎？

另外，在任何人說之前，我知道它的垃圾代碼，但這是我的第一個項目，任何簡化和壓縮它的建議將不勝感激！

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

如果第一個輸入有效，您的代碼工作正常。 但是，您的遞歸調用失敗，因為返回值為 None。 當用戶在輸入后說不並且給出無效值時，會發生遞歸調用。 您遞歸調用 function 但如果選擇主線程，它不會返回選擇的值。 一種解決方案是在主線程上使用條件循環來確保選擇正確的序列。 你可以循環你的 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

我有一個建議，我已經嘗試過你的游戲。 你應該讓它對用戶更友好。 例如

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

，您應該在最后添加可能的答案。 這樣玩起來會容易一些。 例子：

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

我試圖幫助你，但我不明白你的問題和你想要做什么。 請詳細說明。