output - 如果 Python 3.7 中不需要，如何跳过一个条件？

Question

我使用 if、try/except 子句编写了一个代码。我想用“try”来检查参数是否正确，如果正确，“print”函数就会运行。如果参数不正确，则将打印错误消息并且打印部分将不会运行。问题是当我输入正确时它正在运行，但是当我输入错误时，在打印错误消息后我得到 NameError，说“room1”没有定义。我明白为什么会发生这种情况，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)

标签： outputnameerrorpython-3.7

Answer 1

Your print should be in the try segment of your code as it will always execute whether there is an error or not.

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))

Initialization

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