Menu driven program in Python (while loop, shopping list)

A menu driven program shows a list of choices, does what the user picks, and then shows the menu again. A while True loop keeps it running until the user chooses to quit.

The program

python

shopping = []

while True:
    print("Menu: 1 Add  2 Show  3 Remove  4 Quit")
    choice = input("Your choice: ").strip()

    if choice == "1":
        item = input("Item to add: ").strip()
        if item:
            shopping.append(item)
            print(f"Added {item}.")
        else:
            print("Nothing added.")
    elif choice == "2":
        if shopping:
            for number, item in enumerate(shopping, start=1):
                print(f"  {number}. {item}")
        else:
            print("The list is empty.")
    elif choice == "3":
        item = input("Item to remove: ").strip()
        if item in shopping:
            shopping.remove(item)
            print(f"Removed {item}.")
        else:
            print(f"{item} is not on the list.")
    elif choice == "4":
        print("Goodbye.")
        break
    else:
        print("Please choose 1, 2, 3 or 4.")

Typed in this example: 1, milk, 1, bread, 1, eggs, 3, butter, 3, bread, 5, 2, 4. Press Run to type your own.

output

Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 1
Item to add: milk
Added milk.
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 1
Item to add: bread
Added bread.
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 1
Item to add: eggs
Added eggs.
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 3
Item to remove: butter
butter is not on the list.
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 3
Item to remove: bread
Removed bread.
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 5
Please choose 1, 2, 3 or 4.
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 2
  1. milk
  2. eggs
Menu: 1 Add  2 Show  3 Remove  4 Quit
Your choice: 4
Goodbye.

How it works

Each choice is compared as text ("1", not 1). input() always returns a string, so there is no need to convert it, and a typed letter cannot cause a ValueError.
The else at the end catches every answer that is not on the menu, such as 5, and the loop shows the menu again.
break is the only way out of while True. It runs only for choice 4.
shopping.remove(item) raises a ValueError if the item is not in the list, so the program checks item in shopping first.
enumerate(shopping, start=1) numbers the items from 1 instead of 0.

Another way: match

From Python 3.10, a match statement can replace the if/elif chain. Each case is one menu choice, and case _: catches everything else:

python

def handle(choice):
    match choice:
        case "1":
            return "add"
        case "2":
            return "show"
        case "3":
            return "remove"
        case "4":
            return "quit"
        case _:
            return "unknown choice"


for choice in ["2", "4", "9"]:
    print(choice, "->", handle(choice))

output

2 -> show
4 -> quit
9 -> unknown choice

The match statement program shows more of what match can do. The to-do list project builds a bigger menu program that also saves the list to a file.