raw_input每次都返回相同的东西吗？

Question

所以我在一个小文本冒险中为一个商店做了一个函数。代码如下：

def shop():
global p #prompt for raw_input
global gold
global arrows

print"""
You enter the small shop and the man at the counter asks "How can I help?"
"""

print"""
"What would you like to buy?" He asks
You look up at the list of items
Copper Axe - 50G
Copper Sword - 50G
Wooden Bow - 30G
Copper Arrows (20) - 20G
Iron Axe - 100G
Iron Sword - 100G
Iron Arrows (20) - 40G
Steel Axe - 200G
Steel Sword - 200G
Steel Arrows (20) - 80G

or type 'exit' to exit the store
    """
print"You have:"
print inv

op = raw_input(p)

if op == "copper axe" or "Copper axe" or "Copper Axe" or "copper Axe":
    if gold < 50:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 50
        inv.append("Copper Axe")
        shop()
if op == "copper sword" or "copper Sword" or "Copper sword" or "Copper Sword":
    if gold < 50:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 50
        inv.append("Copper Sword")
        shop()
elif op == "Wooden bow" or "wooden bow" or "Wooden Bow" or "wooden Bow":
    if gold < 30:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 30
        inv.append("Wooden Bow")
        if arrows < 10:
            print"You should probably buy some arrows"
            shop()
elif op == "Copper Arrows" or "copper arrows" or "Copper arrows" or "copper Arrows":
    if gold < 20:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 20
        arrows = arrows + 20
        inv.append("Copper Arrows: %p" % arrows)
        shop()
elif op == "iron axe" or "Iron axe" or "Iron Axe" or "iron Axe":
    if gold < 100:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 100
        inv.append("Iron Axe")
        shop()
elif op == "iron sword" or "Iron Sword" or "Iron sword" or "iron Sword":
    if gold < 100:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 100
        inv.append("Iron Sword")
        shop()
elif op == "iron arrows" or "Iron arrows" or "Iron Arrows" or "iron Arrows":
    if gold < 40:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 40
        iarrows = iarrows + 20
        inv.append("Iron arrows %a" % iarrows)
        shop()
elif op == "Steel Axe" or "steel axe" or "Steel axe" or "steel Axe":
    if gold < 200:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 200
        inv.append("Steel Axe")
        shop()
elif op == "Steel Sword" or "steel sword" or "Steel sword" or "steel Sword":
    if gold < 200:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 200
        inv.append("Steel Sword")
        shop()
elif op == "Steel Arrows" or "steel arrows" or "Steel arrows" or "steel Arrows":
    if gold < 80:
        print"You do not have enough gold"
        shop()
    else:
        gold = gold - 80
        sarrows = sarrows + 20
        inv.append("Steel Arrows - %a" % sarrows)
elif op == "exit" or "Exit":
    menu()
else:
    synerr() #syntax error function

基本上，每次我运行它时，无论原始输入是什么，它都会以铜斧头的形式返回。如下所示：

You enter the small shop and the man at the counter asks "How can I help?"


"What would you like to buy?" He asks
You look up at the list of items
Copper Axe - 50G
Copper Sword - 50G
Wooden Bow - 30G
Copper Arrows (20) - 20G
Iron Axe - 100G
Iron Sword - 100G
Iron Arrows (20) - 40G
Steel Axe - 200G
Steel Sword - 200G
Steel Arrows (20) - 80G

or type 'exit' to exit the store

You have:
[]
>>>fkjas;ldkf

You enter the small shop and the man at the counter asks "How can I help?"


"What would you like to buy?" He asks
You look up at the list of items
Copper Axe - 50G
Copper Sword - 50G
Wooden Bow - 30G
Copper Arrows (20) - 20G
Iron Axe - 100G
Iron Sword - 100G
Iron Arrows (20) - 40G
Steel Axe - 200G
Steel Sword - 200G
Steel Arrows (20) - 80G

or type 'exit' to exit the store

You have:
['Copper Axe']
>>>

那么，有谁知道这可能是什么原因？

Answer 1

此表达式：

op == "copper axe" or "Copper axe" or "Copper Axe" or "copper Axe"

是这样评估的：

(op == "copper axe") or "Copper axe" or "Copper Axe" or "copper Axe"

这是你想要的：

op == "copper axe" or op == "Copper axe" or op == "Copper Axe" or op == "copper Axe"

顺便说一句，这里有一个快捷键，转换为小写：

op.lower() == "copper axe"