Not getting expected Output in from conditional statements

Question

print("Pizza price program\n")
size=input("\nWhich size pizza you want? S,L,M\n")
add_pepperoni=input("\nDo you want to add pepperoni to your pizza? Y/N\n ")
extra_cheese=input("\nDo you want to add extra chees on your pizza? Y/N\n")
prize=0
if((size=="S") or (size=="s")):
    prize+=15
elif((size=="M") or (size=="m")):
    prize+=20
elif((size=="L") or (size=="l")):
    prize+=30
else:
    print("\nWrong choice")
if((add_pepperoni=="Y") or (add_pepperoni=="y")):
    if((size=="S") or (size=="s")):
        prize+=2
    elif((size=="M") or (size=="m") or (size=="L") or (size=="l")):
        prize+=3
if((extra_cheese=="Y") or (extra_cheese=="y")):
    prize+=1
        
print(f"\nAmount to be paid for your pizza is {prize}.Rs")

in the first if else block if I provide other than S,L,M the code is proceeding to next if block but I want it to print "Wrong choice" if input is other than S,L,M.

What do you want to do if the size is not S/M/L ? Quit the application? — emmunaf
– emmunaf, Commented Dec 23, 2020 at 15:38

GTBebbo · Accepted Answer · 2020-12-23 15:56:17Z

1

You should use a while loop to continually ask the user until they give a valid choice:


print("Pizza price program\n")

# Initialize the size variable to None
size = None

while size is None:
    # Get user input and make it uppercase so we only have to check for uppercase characters
    user_input = input("Which size pizza you want? S,L,M").upper()
    # If input is in the valid options return it
    if user_input in ['S', 'M', 'L']:
        # If the size is a valid input, set the value
        size = user_input
    else:
        # Otherwise leave the value of size as None and of invalid input
        print("\nWrong choice")

# We now know the size must be one of 'S', 'M' or 'L' so:
prize=0
if size == "S":
    prize+=15
elif size == "M":
    prize+=20
else:  # size must be "L" if it reaches this point so elif is not required
    prize+=30

We can then apply this to the other inputs in your program, wrapping our repeated code in a well-named function:


def get_option(msg, valid_options):
    while True:
        # Get user input, make it uppercase and strip off any whitespace
        user_input = input(msg).upper().strip()
        # If input is in the valid options return it
        if user_input in valid_options:
            return user_input
        else:
            # Otherwise continue to prompt user and warn of invalid input
            print("\nWrong choice")

you can then use this in your code to request the options:


print("Pizza price program\n")
size = get_option("\nWhich size pizza you want? S,L,M\n", ['S', 'M', 'L'])
add_pepperoni = get_option("\nDo you want to add pepperoni to your pizza? Y/N\n ", ['Y', 'N'])
extra_cheese = get_option("\nDo you want to add extra chees on your pizza? Y/N\n", ['Y', 'N'])

# ...

edited Dec 23, 2020 at 15:56

answered Dec 23, 2020 at 15:39

GTBebbo

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9 Comments

Brambor Over a year ago

The OP clearly is learning. You could include an answer without functions. Then, as a bonus, use functions and say "To not duplicate code we can use functions..."

Brambor Over a year ago

Also, size is not a Python function. I found that some size is in NumPy. So I wouldn't warn about function overwrite.

GTBebbo Over a year ago

I noticed @Brambor, however, I thought that would make the code overly complex and harder to understand. This way I highlight the important concept to grasp. Although I appreciate your point of view and will edit to include a version without functions.

Brambor Over a year ago

You don't have to do the whole thing, maybe just one input ;)

GTBebbo Over a year ago

@Brambor, how's that? I've tried to give it a tutorial vibe

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