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How to print an error saying that an integer must be entered if either input value is not a number, and multiply x by y when no error is caught [duplicate]

This question already has answers here:
How can I read inputs as numbers?
(10 answers)
Asking the user for input until they give a valid response
(22 answers)
Closed 9 months ago.
I am trying to make it so that if either input value is not a number, it will print an error message saying that an integer must be entered. And whenever no error is caught, it will multiply x by y. How would I do this?
Here is what I have so far:
def main():
try:
x = int(input("Please enter a number: "))
y = int(input("Please enter another number: "))
except:

It's good practice to not have empty except as it makes debugging difficult
def main():
try:
x = int(input("Please enter a number: "))
y = int(input("Please enter another number: "))
print(x*y)
except ValueError:
print("An integer must be entered. ")
main()

you could try this:
def main():
while True:
try:
x = int(input("Please enter a number: "))
y = int(input("Please enter another number: "))
print(x*y)
break
except:
print("Please input an integer")

while True:
try:
x, y = map(int, input("Please enter two number: ").strip().split())
return x * y
except:
print("two integer must be entered")

Try, except, finally is the most pythonic way of doing this. Note that the code below will only raise an error if the value entered isn't a base 10 integer (which is probably what you want, but worth noting).
try:
x = int(input("Please enter a number: "))
y = int(input("Please enter another number: "))
except ValueError:
print("Entered values must be a base 10 integer")
finally:
print(x*y)

'Perfect factorial' python program [duplicate]

This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 8 years ago.
I need to check whether what the user entered is positive. If it is not I need to print an error in the form of a msgbox.
number = input("Enter a number: ")
###################################
try:
val = int(number)
except ValueError:
print("That's not an int!")
The above code doesn't seem to be working.
Any ideas?

while True:
number = input("Enter a number: ")
try:
val = int(number)
if val < 0: # if not a positive int print message and ask for input again
print("Sorry, input must be a positive integer, try again")
continue
break
except ValueError:
print("That's not an int!")
# else all is good, val is >= 0 and an integer
print(val)

what you need is something like this:
goodinput = False
while not goodinput:
try:
number = int(input('Enter a number: '))
if number > 0:
goodinput = True
print("that's a good number. Well done!")
else:
print("that's not a positive number. Try again: ")
except ValueError:
print("that's not an integer. Try again: ")
a while loop so code continues repeating until valid answer is given, and tests for the right input inside it.

How to stop a while loop processing integers with a string

I'm processing integer inputs from a user and would like for the user to signal that they are done with inputs by typing in 'q' to show they are completed with their inputs.
Here's my code so far:
(Still very much a beginner so don't flame me too much)
def main():
print("Please enter some numbers. Type 'q' to quit.")
count = 0
total = 0
num=[]
num = int(input("Enter a number: "))
while num != "q":
num = int(input("Enter a number: "))
count += 1
total += num
del record[-1]
print (num)
print("The average of the numbers is", total / count)
main()
Any feedback is helpful!

You probably get a ValueError when you run that code. That's python's way of telling you that you've fed a value into a function that can't handle that type of value.
Check the Exceptions docs for more details.
In this case, you're trying to feed the letter "q" into a function that is expecting int() on line 6. Think of int() as a machine that only handles integers. You just tried to stick a letter into a machine that's not equipped to handle letters and instead of bursting into flames, it's rejecting your input and putting on the breaks.
You'll probably want to wrap your conversion from str to int in a try: statement to handle the exception.
def main():
num = None
while num != "q":
num = input("Enter number: ")
# try handles exceptions and does something with them
try:
num = int(num)
# if an exception of "ValueError" happens, print out a warning and keep going
except ValueError as e:
print(f'that was not a number: {e}')
pass
# if num looks like an integer
if isinstance (num, (int, float)):
print('got a number')
Test:
Enter number: 1
got a number
Enter number: 2
got a number
Enter number: alligator
that was not a number: invalid literal for int() with base 10: 'alligator'
Enter number: -10
got a number
Enter number: 45
got a number
Enter number: 6.0222
that was not a number: invalid literal for int() with base 10: '6.0222'
I'll leave it to you to figure out why 6.02222 "was not a number."

changing your code as little as possible, you should have...
def main():
print("Please enter some numbers. Type 'q' to quit.")
count = 0
total = 0
num=[]
num.append(input("Enter a number: "))
while num[-1] != "q":
num.append(input("Enter a number: "))
count += 1
try:
total += int(num[-1])
except ValueError as e:
print('input not an integer')
break
print (num)
print("The average of the numbers is", total / count)
main()

You may attempt it the way:
def main():
num_list = [] #list for holding in user inputs
while True:
my_num = input("Please enter some numbers. Type 'q' to quit.")
if my_num != 'q':
num_list.append(int(my_num)) #add user input as integer to holding list as long as it is not 'q'
else: #if user enters 'q'
print(f'The Average of {num_list} is {sum(num_list)/len(num_list)}') #calculate the average of entered numbers by divide the sum of all list elements by the length of list and display the result
break #terminate loop
main()

Why is the error raised regardless of input? [duplicate]

This question already has answers here:
How can I check if string input is a number?
(30 answers)
Closed 3 years ago.
I am working my way through Python for Everyone and I am stuck at this junction. To my eye I have stated that the ValueError is only to be raised if 'num' is anything other than a integer. However when I run the code the error is raised everytime regardless of input. Can anyone nudge me in the right direction?
Extensively googled but I'm not entirely too sure what specifically I should google for...
largest = None
smallest = None
while True:
try:
num = input("Enter a number: ")
if num != int : raise ValueError
elif num == "done" : break
except ValueError:
print("Error. Please enter an integer or type 'done' to run the program.")
quit()
print("Maximum", largest)
print("Minimum", smallest)
The code always raises ValueError even when the input is an integer.

This line checks if the inputted string is literally equal to the builtin type int:
if num != int : raise ValueError
Other problem is that the input() function always returns a string. So if you want to raise a ValueError when the user inputs anything but a number, simply do:
inputted = input("Enter a number: ")
num = int(inputted) # raises ValueError when cannot be converted to int

If you want to check if the string entered can be converted into an int, just try it:
while True:
num = input("Enter a number: ")
if num == "done":
break
try:
num = int(num)
except ValueError:
continue

Python input validation: how to limit user input to a specific range of integers? [duplicate]

This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 1 year ago.
Beginner here, looking for info on input validation.
I want the user to input two values, one has to be an integer greater than zero, the next an integer between 1-10. I've seen a lot of input validation functions that seem over complicated for these two simple cases, can anyone help?
For the first number (integer greater than 0, I have):
while True:
try:
number1 = int(input('Number1: '))
except ValueError:
print("Not an integer! Please enter an integer.")
continue
else:
break
This also doesn't check if it's positive, which I would like it to do. And I haven't got anything for the second one yet. Any help appreciated!

You could add in a simple if statement and raise an Error if the number isn't within the range you're expecting
while True:
try:
number1 = int(input('Number1: '))
if number1 < 1 or number1 > 10:
raise ValueError #this will send it to the print message and back to the input option
break
except ValueError:
print("Invalid integer. The number must be in the range of 1-10.")

Use assert:
while True:
try:
number1 = int(input('Number1: '))
assert 0 < number1 < 10
except ValueError:
print("Not an integer! Please enter an integer.")
except AssertionError:
print("Please enter an integer between 1 and 10")
else:
break

class CustomError(Exception):
pass
while True:
try:
number1 = int(raw_input('Number1: '))
if number1 not in range(0,9):
raise CustomError
break
except ValueError:
print("Numbers only!")
except CustomError:
print("Enter a number between 1-10!)