PYTHON· LESSON 05

If Else and Menu Driven Programs

Learn conditions, comparisons, if/elif/else statements, and how to build menu-driven programs.

Programs often need to make decisions. For example, a program may check whether OSDC registrations are open, decide whether a workshop has available seats, or perform an action selected from a menu.

Python uses conditional statements to execute different code based on whether a condition is true or false.

Conditions

A condition is an expression that evaluates to either True or False.

member_count = 50

print(member_count > 0)
print(member_count == 50)
print(member_count < 10)

Comparison operators:

Operator	Meaning	Example
`==`	Equal to	`count == 50`
`!=`	Not equal to	`count != 50`
`>`	Greater than	`count > 50`
`<`	Less than	`count < 50`
`>=`	Greater than or equal to	`count >= 50`
`<=`	Less than or equal to	`count <= 50`

Do not confuse = and ==:

member_count = 50          # assignment
print(member_count == 50)  # comparison

The `if` Statement

The if statement runs a block of code only when its condition is true.

registration_open = True

if registration_open:
    print("OSDC registrations are open.")

Python uses indentation to define the block belonging to an if statement. The usual indentation is four spaces, and the colon after the condition is required.

member_count = 50

if member_count > 0:
    print("OSDC has registered members.")

Using Input with `if`

input() returns a string, so convert numeric input before comparing it with a number.

seat_count = int(input("Enter the number of available seats: "))

if seat_count > 0:
    print("Seats are available.")

For text, normalize input before comparing it:

technology = input("Enter a technology: ").strip().lower()

if technology == "python":
    print("Python is used in the backend workshop.")

The `if-else` Statement

Use else when one block should run if the if condition is false.

registration_open = False

if registration_open:
    print("You can register for the workshop.")
else:
    print("Registration is currently closed.")

Only one of the two blocks runs.

member_count = int(input("Enter the number of OSDC members: "))

if member_count >= 10:
    print("The club has a large member group.")
else:
    print("The club has a small member group.")

The `if-elif-else` Statement

Use elif, short for else if, when there are multiple possible conditions.

attendance = int(input("Enter workshop attendance: "))

if attendance >= 50:
    print("Excellent attendance.")
elif attendance >= 25:
    print("Good attendance.")
elif attendance > 0:
    print("Some members attended.")
else:
    print("No attendance recorded.")

Python checks conditions from top to bottom. Once a condition is true, its block runs and the remaining conditions are skipped.

The order of conditions matters:

score = 85

if score >= 80:
    print("Excellent score.")
elif score >= 40:
    print("Passed.")
else:
    print("Did not pass.")

An else block is optional. An if statement can have multiple elif blocks, but only one else block.

Logical Operators

Logical operators combine conditions.

`and`

and is true only when both conditions are true.

member_count = 50
registration_open = True

if member_count > 0 and registration_open:
    print("OSDC can accept registrations.")

`or`

or is true when at least one condition is true.

technology = input("Enter HTML, CSS, or Python: ").strip().lower()

if technology == "html" or technology == "css":
    print("This is a frontend technology.")

For multiple possible values, membership testing is often clearer:

technology = input("Enter a technology: ").strip().lower()

if technology in ("html", "css", "javascript"):
    print("Frontend technology selected.")

`not`

not reverses a Boolean value.

registration_open = False

if not registration_open:
    print("Registrations are closed.")

Use parentheses to make complex conditions clear:

member_count = 30
registration_open = True

if (member_count > 0 and registration_open) or member_count == 0:
    print("The registration system can be displayed.")

Truthy and Falsy Values

Python allows many values to be used directly as conditions.

The following values are falsy:

False
None
0
0.0
""
Empty collections such as [], {}, (), and set()

Most other values are truthy.

club_name = input("Enter the club name: ").strip()

if club_name:
    print(f"Welcome to {club_name}.")
else:
    print("The club name cannot be empty.")

Nested Conditions

An if statement can be placed inside another if statement.

registration_open = True
seat_count = 20

if registration_open:
    if seat_count > 0:
        print("Registration is available.")
    else:
        print("The workshop is full.")
else:
    print("Registration is closed.")

The same logic can often be expressed more simply:

if registration_open and seat_count > 0:
    print("Registration is available.")
else:
    print("Registration is unavailable.")

Conditional Expressions

A conditional expression selects one of two values in a single line.

seat_count = 10
status = "Seats available" if seat_count > 0 else "Workshop full"

print(status)

The general syntax is:

value_if_true if condition else value_if_false

Use a regular if-else statement when the logic contains multiple steps. Conditional expressions are best for short assignments or output values.

Validating Conditions

Conditions are useful for validating user input.

try:
    seat_count = int(input("Enter the number of workshop seats: "))

    if seat_count <= 0:
        print("The number of seats must be greater than zero.")
    else:
        print(f"The workshop has {seat_count} seats.")
except ValueError:
    print("Please enter a whole number.")

A range can be validated using chained comparisons:

rating = float(input("Enter the workshop rating from 0 to 5: "))

if 0 <= rating <= 5:
    print("Valid rating.")
else:
    print("Rating must be between 0 and 5.")

A menu-driven program displays a list of choices and performs an action based on the user’s selection.

A basic menu has three parts:

Display the available options.
Read the user’s choice.
Use conditional logic to perform the selected action.

print("OSDC Workshop Menu")
print("1. View workshops")
print("2. View club information")
print("3. Exit")

choice = input("Enter your choice: ").strip()

if choice == "1":
    print("Available workshops: HTML, CSS, JavaScript, Python, FastAPI")
elif choice == "2":
    print("OSDC is the Open Source Developers Community at JIIT, Noida.")
elif choice == "3":
    print("Goodbye.")
else:
    print("Invalid choice.")

The menu choice is read as a string, so compare it with values such as "1" and "2".

A while loop keeps the menu available until the user chooses to exit.

while True:
    print("\nOSDC Workshop Menu")
    print("1. View workshops")
    print("2. Register for a workshop")
    print("3. View club information")
    print("4. Exit")

    choice = input("Enter your choice: ").strip()

    if choice == "1":
        print("Available workshops: HTML, CSS, JavaScript, Python, FastAPI")
    elif choice == "2":
        print("Workshop registration selected.")
    elif choice == "3":
        print("OSDC is the Open Source Developers Community at JIIT, Noida.")
    elif choice == "4":
        print("Thank you for using the OSDC menu.")
        break
    else:
        print("Invalid choice. Please select an option from 1 to 4.")

break immediately stops the loop when the user selects the exit option.

Functions keep each menu action separate and make the program easier to maintain.

def show_workshops():
    print("Available workshops:")
    print("- HTML and CSS")
    print("- JavaScript")
    print("- Python and FastAPI")


def show_club_information():
    print("OSDC: Open Source Developers Community")
    print("Institution: JIIT, Noida")


def show_menu():
    print("\nOSDC Workshop Menu")
    print("1. View workshops")
    print("2. View club information")
    print("3. Exit")


while True:
    show_menu()
    choice = input("Enter your choice: ").strip()

    if choice == "1":
        show_workshops()
    elif choice == "2":
        show_club_information()
    elif choice == "3":
        print("Goodbye.")
        break
    else:
        print("Invalid choice.")

print("Calculator Menu")
print("1. Add")
print("2. Subtract")
print("3. Multiply")
print("4. Divide")

choice = input("Choose an operation: ").strip()

if choice in ("1", "2", "3", "4"):
    first_number = float(input("Enter the first number: "))
    second_number = float(input("Enter the second number: "))

    if choice == "1":
        print("Result:", first_number + second_number)
    elif choice == "2":
        print("Result:", first_number - second_number)
    elif choice == "3":
        print("Result:", first_number * second_number)
    elif choice == "4":
        if second_number == 0:
            print("A number cannot be divided by zero.")
        else:
            print("Result:", first_number / second_number)
else:
    print("Invalid operation.")

The outer condition checks whether the menu choice is valid. The inner condition prevents division by zero.

def show_workshops():
    print("\nAvailable workshops")
    print("- HTML and CSS")
    print("- JavaScript")
    print("- Python")
    print("- FastAPI")


def register_for_workshop():
    workshop = input("Enter the workshop name: ").strip()

    if not workshop:
        print("Workshop name cannot be empty.")
    else:
        print(f"Registration request received for {workshop}.")


def show_club_information():
    print("\nOSDC")
    print("Open Source Developers Community")
    print("JIIT, Noida")


while True:
    print("\nOSDC Full Stack Workshop")
    print("1. View workshops")
    print("2. Register for a workshop")
    print("3. View club information")
    print("4. Exit")

    choice = input("Enter your choice: ").strip()

    if choice == "1":
        show_workshops()
    elif choice == "2":
        register_for_workshop()
    elif choice == "3":
        show_club_information()
    elif choice == "4":
        print("Thank you for attending the OSDC workshop.")
        break
    else:
        print("Invalid choice. Please select 1, 2, 3, or 4.")

A menu should reject invalid choices instead of crashing or performing an unintended action.

while True:
    choice = input("Choose 1, 2, or 3: ").strip()

    if choice in ("1", "2", "3"):
        break

    print("Invalid choice. Please try again.")

print(f"You selected option {choice}.")

For a numeric menu, validate the conversion as well:

while True:
    try:
        choice = int(input("Choose 1, 2, or 3: "))

        if choice in (1, 2, 3):
            break

        print("Please choose a number from 1 to 3.")
    except ValueError:
        print("Please enter a number.")

print(f"You selected option {choice}.")

`match-case` for Menus

Python 3.10 and later provide match-case, which is useful when one value must be compared with several fixed patterns.

choice = input("Enter 1, 2, or 3: ").strip()

match choice:
    case "1":
        print("View workshops selected.")
    case "2":
        print("View club information selected.")
    case "3":
        print("Exit selected.")
    case _:
        print("Invalid choice.")

The underscore _ is the default case. It matches anything that did not match an earlier case.

Use if-elif-else when conditions involve ranges or complex expressions. Use match-case when comparing one value with several known patterns.

Common Mistakes

Using Assignment Instead of Comparison

choice = input("Enter a choice: ")

# if choice = "1":  # SyntaxError
if choice == "1":
    print("Option 1 selected.")

Forgetting the Colon

# if choice == "1"
#     print("Option 1 selected.")

if choice == "1":
    print("Option 1 selected.")

Comparing Input with an Integer Without Conversion

choice = input("Enter 1 or 2: ")

if choice == 1:
    print("Option 1 selected.")

The comparison fails because choice is a string. Either compare with "1" or convert the input:

choice = int(input("Enter 1 or 2: "))

if choice == 1:
    print("Option 1 selected.")

Forgetting to Handle Invalid Choices

Every menu should have an else branch or another way to report an invalid choice.

choice = input("Enter a menu choice: ").strip()

if choice == "1":
    print("Option 1 selected.")
elif choice == "2":
    print("Option 2 selected.")
else:
    print("Invalid choice.")

Always provide a clear exit condition when using while True for a menu.

while True:
    choice = input("Enter 1 to exit: ").strip()

    if choice == "1":
        print("Exiting.")
        break

Quick Reference

if condition:
    pass

if condition:
    pass
else:
    pass

if first_condition:
    pass
elif second_condition:
    pass
else:
    pass

if condition_a and condition_b:
    pass

if condition_a or condition_b:
    pass

if not condition:
    pass

while True:
    choice = input("Choose an option: ")

    if choice == "1":
        pass
    elif choice == "2":
        pass
    elif choice == "3":
        break
    else:
        print("Invalid choice.")

If Else and Menu Driven Programs

Conditions

The `if` Statement

Using Input with `if`

The `if-else` Statement

The `if-elif-else` Statement

Logical Operators

`and`

`or`

`not`

Truthy and Falsy Values

Nested Conditions

Conditional Expressions

Validating Conditions

Menu-Driven Programs

Basic Menu

Repeating a Menu

Menu with Functions

Menu-Driven Calculator

Complete OSDC Workshop Menu

Menu Choice Validation

`match-case` for Menus

Common Mistakes

Using Assignment Instead of Comparison

Forgetting the Colon

Comparing Input with an Integer Without Conversion

Forgetting to Handle Invalid Choices

Forgetting to Exit a Menu Loop

Quick Reference

If Else and Menu Driven Programs

Conditions

The if Statement

Using Input with if

The if-else Statement

The if-elif-else Statement

Logical Operators

and

or

not

Truthy and Falsy Values

Nested Conditions

Conditional Expressions

Validating Conditions

Menu-Driven Programs

Basic Menu

Repeating a Menu

Menu with Functions

Menu-Driven Calculator

Complete OSDC Workshop Menu

Menu Choice Validation

match-case for Menus

Common Mistakes

Using Assignment Instead of Comparison

Forgetting the Colon

Comparing Input with an Integer Without Conversion

Forgetting to Handle Invalid Choices

Forgetting to Exit a Menu Loop

Quick Reference

The `if` Statement

Using Input with `if`

The `if-else` Statement

The `if-elif-else` Statement

`and`

`or`

`not`

`match-case` for Menus