如果 Python 3.7 不需要，如何跳过一个条件？

Question

我使用 if, try/except 子句编写了代码。我想使用 "try" 检查参数是否正确，如果正确，"print" 函数将运行。如果参数不对则打印错误信息并且打印部分不会运行。问题是当我输入正确时它是运行ning 但是当我输入错误时，打印错误消息后我得到 NameError，说 "room1" 未定义。我明白为什么会这样，但我很困惑如何在不出错的情况下获得正确的输出。

我的代码是：

class Hotel:
    def __init__(self,room,catagory):
        if type(room) != int:
            raise TypeError()
        if type(catagory) != str:
            raise TypeError()
        self.room = room
        self.catagory = catagory
        self.catagories = {"A":"Elite","B":"Economy","C":"Regular"}
        self.rooms = ["0","1","2","3","4","5"]


    def getRoom(self):
        return self.room

    def getCatagory(self):

        return self.catagories.get(self.catagory)
    def __str__(self):
        return "%s and %s"%(self.rooms[self.room],self.catagories.get(self.catagory))

try:
    room1 = Hotel(a,"A")

except: 

    print("there's an error")


print (room1)

Answer 1

您的打印应该在代码的 try 段中，因为无论是否有错误，它都会始终执行。

class Hotel:
    def __init__(self,room,catagory):
        if type(room) != int:
            raise TypeError()
        if type(catagory) != str:
            raise TypeError()
        self.room = room
        self.catagory = catagory
        self.catagories = {"A":"Elite","B":"Economy","C":"Regular"}
        self.rooms = ["0","1","2","3","4","5"]


    def getRoom(self):
        return self.room

    def getCatagory(self):

        return self.catagories.get(self.catagory)
    def __str__(self):
        return "%s and %s"%(self.rooms[self.room],self.catagories.get(self.catagory))

初始化

try:
    room1 = Hotel(a,"A")
    print (room1)
except: 
    print("there's an error")

如果 Python 3.7 不需要，如何跳过一个条件？

How to skip one condition if that is not needed in Python 3.7?

nameerror

output

python-3.7