How do you execute different options based on how many times a function is run?

Question

Hello I am new to python and this is my first stack overflow post, so I apologize if this question has been answered elsewhere (I'm not entirely sure how to search for this topic, feel free to drop a link if it is).

I'm creating an adventure game and I want the user to choose different paths within a house.

I want the user to be able to travel back and forth between room A and room B, however I want different dialogue to appear based on how many times a user has entered room A or room B. If its the first time, I want the function to print (option 1), but if the user enters >1 time i want the function to print (option 2)

I've figured out how to travel back and forth between the 2 rooms, but I don't get different dialogue options if I enter either room >1 times with my current increment code. Does it have to do with the fact that I exit the room_A function when i 'travel to room_B'?

Any help would be greatly appreciated!

def room_A():
     room_A_count = 0

# - if this is the first time the user enters the room, this code runs...
     if room_A_count == 0:
          print ("You enter room A")
          print ("What would you like to do?")
          print ("1. Enter room B")
          print ("2. nothing")
          room_A_choice = input("Choice: ")

          if room_A_choice = "1":
               room_A_count += 1
               # - travel to room B
               room_B()
          elif room_A_choice == "2":
               print ("You do nothing")

# - if the user travels back from Room B to Room A, it prints this message...
     else: 
          print ("You enter room A again") 

Answer 1

One thing I'd say is that a good principle to generally follow is for functions to behave the same way every time you call them. With that being said, one work around could be to store a room_A_count as a global variable. Then pass it into the function as an argument and increment each time.

That's a bit hacky and seems like it could break easy, so I'd propose an object oriented solution-

class RoomA:
    def __init__(self):
        self.room_A_count = 0

    def enter(self):
        if self.room_A_count == 0:
          print ("You enter room A")
          print ("What would you like to do?")
          print ("1. Enter room B")
          print ("2. nothing")
          room_A_choice = input("Choice: ")

          if room_A_choice = "1":
               self.room_A_count += 1
               # - travel to room B
               room_B()
          elif room_A_choice == "2":
               print ("You do nothing")

        else: 
            print ("You enter room A again")

Answer 2

I'm unable to comment, so I apologize if I understand your question right, all you'll need to do is add an if statement where it checks for room_A_choice

          if room_A_choice = ="1":
               room_A_count += 1
               if room_A_count >= 2:
                    # do something
               else : room_B()

Answer 3

You can use function static variables

def room_A():
  if not hasattr(room_A, 'room_A_count'): # check if static variable exists
    room_A.room_A_count = 0               # initialize if doesn't set

  if room_A.room_A_count == 0:  # refers to the funtion static variable
    print ("You enter room A")
    print ("What would you like to do?")
    print ("1. Enter room B")
    print ("2. nothing")
    room_A_choice = input("Choice: ")

    if room_A_choice == "1":
          room_A.room_A_count += 1 # incremention function static variable
          # - travel to room B
          room_B()
    elif room_A_choice == "2":
          print ("You do nothing")

    # - if the user travels back from Room B to Room A, it prints this message...
  else: 
    print ("You enter room A again") 

def room_B():
  print('In room B')

room_A()                 # First call
room_A()                 # Second call

Output

You enter room A
What would you like to do?
1. Enter room B
2. nothing
Choice: 1
In room B
You enter room A again