Python如果语句没有正确读取

Question

所以我正在开发一个基于文本的游戏,我有一个商店,你可以卖出矿石.我有一个if语句,检查您要出售的矿石数量是否大于您拥有的数量.无论我输入什么号码,我都说不够.

copore = 100 #testing
    op = raw_input(p)
    if op.lower() == "copper":
        print"""
You have %r copper ore.
How much would you like to sell?
        """ % copore
        op = raw_input(p)
        if op.lower() > copore:
            print"""
You do not have that much copper ore.
You have %r copper ore.
            """ % copore
            menu()
        elif op.lower() <= copore:
            copore = copore - op
            gold = gold + (op * 2)

            print"""
You sell your copper ore for 2 gold each.
You now have %r copper ore.
and %r gold.
            """ % (copore, gold)
            menu()
        else:
            print"""
That is not a valid number
            """
            menu()

这可能很容易成为我自己的愚蠢,但我认为让别人去看它不会有什么坏处.谢谢.

Answer 1

您需要将输入转换为raw_input整数.如果不转换它,你就会进行字符串和整数比较.

>>> a = raw_input("Test amount:")
Test amount:10
>>> type(a)
<type 'str'>
>>> a > 10
True
>>> a > 100
True
>>> b = int(a)
>>> b > 100
False