I am trying to make a simple calculator but i get an error after the input. Please help

Question

I am trying to make a simple calculator but i get an error after the input. Please help!!!

this is the code:

while True:
print("Options:")
print("Type Add to add values")
print("Type Subtract to subtract values")
print("Type Multiply to multiply values")
print("Type Divide to dividde values")
print("Type Quit to exit the calculator")
print("Choose")
user_input = "Quit"
user_input1 = "Add"
user_input2 = "Subtract"
user_input3 = "Multiply"
user_input4 = "Divide"

if user_input == "Quit":
break
elif user_input1 == "Add":
num1 = float(input("Enter a value"))
num2 = float(input("Enter second value"))
result = str(num1 + num2)
print ("The answer is " + result)
elif user_input2 == "Subtract":
num1 = float(input("Enter a value"))
num2 = float(input("Enter second value"))
result = str(num1 + num2)
print ("The answer is " + result)
elif user_input3 == "Multiply":
num1 = float(input("Enter a value"))
num2 = float(input("Enter a ssecond value"))
result = str(num1 + num2)
print("The answer is " + result)
elif user_input4 == "Divide":
num1 = float(input("Enter a value"))
num2 = float(input("Enter another value"))
result = str(num1 + num2)
print("The answer is " + result)

Answer 1

You can use only one variable to store the user input for operation, and have it receive the input as well:

user_input = input("Choose: ")

Also, keep in mind that as far as Python is concerned, 'Add' ! = 'add', so even if your user type 'add', it will not pass the adding condition.
A nice workaround for this is to have

user_input = user_input.lower()

That will convert the user input to lower case, and then you adapt your conditions to accept lower case.

if user_input == "quit":

This way, the program will accept 'quit', 'Quit', 'QUIT', 'QuIT', and so on.

Answer 2

while True:
print("Options:")
print("Type Add to add values")
print("Type Subtract to subtract values")
print("Type Multiply to multiply values")
print("Type Divide to dividde values")
print("Type Quit to exit the calculator")
user_input = input("Choose:")

user_input.lower()

if user_input == "Quit":
break
elif user_input == "Add":
num1 = float(input("Enter a value"))
num2 = float(input("Enter second value"))
result = str(num1 + num2)
print ("The answer is " + result)

elif user_input == "Subtract":
num1 = float(input("Enter a value"))
num2 = float(input("Enter second value"))
result = str(num1 + num2)
print ("The answer is " + result)

elif user_input == "Multiply":
num1 = float(input("Enter a value"))
num2 = float(input("Enter a ssecond value"))
result = str(num1 + num2)
print("The answer is " + result)

elif user_input == "Divide":
num1 = float(input("Enter a value"))
num2 = float(input("Enter another value"))
result = str(num1 + num2)
print("The answer is " + result)
break

try this and lets see

Answer 3

I get your very same error if I use Python 2.7 instead of Python 3.X.
So, which version of Python are you using?

Answer 4

while True:
print("Options:")
print("Type Add to add values")
print("Type Subtract to subtract values")
print("Type Multiply to multiply values")
print("Type Divide to dividde values")
print("Type Quit to exit the calculator")
user_input = input("Choose: ")

user_input = user_input.lower()

if user_input == "quit":
break
elif user_input == "add":
num1 = float(input("Enter a value"))
num2 = float(input("Enter second value"))
result = str(num1 + num2)
print ("The answer is " + result)
elif user_input == "subtract":
num1 = float(input("Enter a value"))
num2 = float(input("Enter second value"))
result = str(num1 + num2)
print ("The answer is " + result)
elif user_input == "multiply":
num1 = float(input("Enter a value"))
num2 = float(input("Enter a ssecond value"))
result = str(num1 + num2)
print("The answer is " + result)
elif user_input == "divide":
num1 = float(input("Enter a value"))
num2 = float(input("Enter another value"))
result = str(num1 + num2)
print("The answer is " + result)

The above runs fine for me on IDLE. Compare it with what you have.

Answer 5

That must be it. The code I provided must be used in Python 3.

Answer 6

it returns this even after the correction!

Traceback (most recent call last):
File "C:/Users/hp/Desktop/My simple Calculator", line 9, in <module>
user_input = input("Choose: ")
File "<string>", line 1, in <module>
NameError: name 'Add' is not defined

Answer 7

I use 2.7 python