python - 使用带有原始输入的函数

Question

我正在尝试到目前为止所学到的知识，我想使用 raw_input() 创建一些交互式的东西。

我想做的是创建一个函数，该函数将创建一个对话，该对话将根据输入走向不同的方向。但是，我无法弄清楚如何使函数接受 raw_input 作为其参数。

这是我编写的代码；

drink = raw_input("Coffee or Tea?")

def drinktype(drink):
    if drink == "Coffee":
    #I WANT TO INSERT A CODE HERE THAT WILL CALL THE FUNCTION coffee(x)
    elif drink == "Tea":
        print "Here is your tea."
    else:
        print "Sorry."

x = raw_input("Americano or Latte?")

def coffee(x):
    if x == "Americano":
        return "Here it is."    
    elif x == "Latte":
        return "Here is your latte."
    else:
        return "We do not have that, sorry."

Answer 1

只有在需要咖啡时，您才需要请求美式咖啡或拿铁咖啡；用户是否需要茶是无关紧要的。一旦将其移至 Coffee 案例下，您只需将返回值传递给对 coffee() 的调用即可。返回值也需要打印。

def drinktype(drink):
    if drink == "Coffee":
        kind = raw_input("Americano or Latte?")
        print coffee(kind)
    elif drink == "Tea":
        print "Here is your tea."
    else:
        print "Sorry."

def coffee(x)
    if x == "Americano":
        return "Here it is."    
    elif x == "Latte":
        return "Here is your latte."
    else:
        return "We do not have that, sorry."

drink = raw_input("Coffee or Tea?")
drinktype(drink)