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

Question

I have written a code using if, try/except clause. I want to use "try" to check whether the parameters are correct and if those are correct the "print" function will run.
If the parameters are not right then the error message will be printed and the print section will not run.
The problem is when I am correct input it is running but when I am giving wrong input, after printing the error message I am getting NameError, saying "room1" is not defined. I understood why it is happening but I am confused how to get the correct output without getting an error.
My code is:
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

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")