我正在尝试到目前为止我学到的东西,我想用object
创建一些互动的东西。
我想要做的是创建一个能够根据输入创建不同方向的对话的功能。但是,我无法弄清楚如何使函数接受raw_input()
作为其参数。
以下是我编写的代码;
raw_input
答案 0 :(得分:1)
如果要求咖啡,您只需要做美式咖啡或拿铁咖啡;如果用户请求茶是无关紧要的。一旦在Coffee案例下移动,您只需将返回的值传递给coffee()
的调用。还需要打印返回值。
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)
答案 1 :(得分:0)
你在找这样的东西吗?
drink = raw_input("Coffee or Tea?")
def drinktype(drink):
if drink == "Coffee":
usercoffeetype = raw_input("What type of coffee do you drink?")
coffee(usercoffeetype)
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."
同样只是一个注释 - 使用像“x”这样的变量名称通常不是一个好主意。如果你的变量名实际上描述了它的含义,那就更好了,比如:
def coffee(coffeechoice)
if coffeechoice == "Americano":
return "Here it is."
elif coffeechoice == "Latte":
return "Here is your latte."
else:
return "We do not have that, sorry."
答案 2 :(得分:0)
看起来你正试图获得类似下面的内容
def coffee():
x = raw_input("Americano or Latte?")
if x == "Americano":
return "Here it is."
elif x == "Latte":
return "Here is your latte."
else:
return "We do not have that, sorry."
def drinktype(drink):
if drink == "Coffee":
print coffee()
elif drink == "Tea":
print "Here is your tea."
else:
print "Sorry."
drink = raw_input("Coffee or Tea?")
drinktype(drink)
请注意 1.正确的缩进对于您的代码工作至关重要 2.在定义函数drinktype()之后,你需要实际调用它来让它运行。 (最后一行是调用函数)