I need to be able to prompt the user to enter an empty string so it can check if the answer is correct. but every time I do that I can error saying invalid literal for int()
so I need to change my user_input so it can accept int() and strings(). how do I make that possible ?
# program greeting
print("The purpose of this exercise is to enter a number of coin values")
print("that add up to a displayed target value.\n")
print("Enter coins values as 1-penny, 5-nickel, 10-dime,and 25-quarter.")
print("Hit return after the last entered coin value.")
print("--------------------")
#print("Enter coins that add up to 81 cents, one per line.")
import sgenrand
#prompt the user to start entering coin values that add up to 81
while True:
total = 0
final_coin= sgenrand.randint(1,99)
print ("Enter coins that add up to", final_coin, "cents, on per line")
user_input = int(input("Enter first coin: "))
if user_input != 1 and user_input!=5 and user_input!=10 and user_input!=25:
print("invalid input")
else:
total = total + user_input
while total <= final_coin:
user_input = int(input("Enter next coin:"))
if user_input != 1 and user_input!=5 and user_input!=10 and user_input!=25:
print("invalid input")
else:
total = total + user_input
if total > final_coin :
print("Sorry - total amount exceeds", (final_coin))
elif total < final_coin:
print("Sorry - you only entered",(total))
else:
print("correct")
goagain= input("Try again (y/n)?:")
if goagain == "y":
continue
elif goagain == "n":
print("Thanks for playing ... goodbye!" )
break
Store the value returned by input() in a variable.
Check that the string is not empty before calling int().
if it's zero, that's the empty string.
otherwise, try int()ing it.
Related
def set_number():
import random
return random.randint(1,500)
#This function plays the game
def number_guessing_game(number):
guess_counter = 0
guess = int(input("Enter a number between 1 and 500."))
while guess != number:
guess_counter += 1
if guess > number:
print(f"You guessed too high. Try Again!")
guess = int(input("Enter a number between 1 and 500."))
elif guess < number:
print(f"You guessed too low. Try Again!")
guess = int(input("Enter a number between 1 and 500."))
if guess == number:
print(f"You guessed the number! Good Job.!")
again = str(input("would you like to play again? Enter 'y' for yes or 'n' to close the game."))
def main():
print(f"Welcome to the Number Guessing Game!\n" +
f"You will have unlimited guesses. The number is between 1 and 500.\n" +
f"Good Luck!")
number = set_number()
guess_count = number_guessing_game(number)
main()
I am working on a simple game project for my coding class. I am not good at coding at all. I came up with this part of the program, I just cannot figure out how to loop the entire number_guessing_game function until the user enters 'n' to stop it, I can't use a break because we did not learn it in the class and I will receive a 0 if I use a break.
I tried nesting a while loop inside of the function but I know I did it wrong.
Instead of using break use return.
def main():
print(f"Welcome to the Number Guessing Game!\n" +
f"You will have unlimited guesses. The number is between 1 and 500.\n" +
f"Good Luck!")
while True:
number = set_number()
number_guessing_game(number)
again = input("would you like to play again? Enter 'y' for yes or 'n' to close the game.")
if again == 'n':
return
main()
You will probably want to remove the last line of the number_guessing_game function if you use this approach
First, your code is assuming the return of input is an integer that can be converted with int(). If you were to give it 'n' your program will crash.
Instead you could use the string class method isdigit() to see if the input was an integer value and then make a logical decision about it. Also note you do not need to convert the return from input to a str() as it is already a str type. You can confirm this with a print(type(input("give me something")))
guess = input("Enter a number between 1 and 500. 'n' to quit"))
if guess.isdigit():
[your code to check the value]
elif ('n' == guess):
return
else:
print(f"You entered an invalid entry: {guess}. Please re-enter a valid value")
If you dont like the idea of using 'return' you could change your while loop to look something like:
while(('n' != guess) or (guess != number)):
If you want the function body looping continuously you could have some code like:
def number_guessing_game(number):
exit_game = False
guess_counter = 0
while(exit_game != True):
guess = input("Enter a number between 1 and 500.))
guess_counter += 1
if guess.isdigit():
if int(guess) > number:
print("You guessed too high. Try Again!")
elif int(guess) < number:
print("You guessed too low. Try Again!")
elif int(guess) == number:
print("You guessed the number! Good Job.!")
again = input("would you like to play again? Enter 'y' for yes or 'n' to close)
if ('n' == again):
exit_game = True
else:
print("Error, please enter a valid value")
def itemPrices():
items = []
while True:
itemAmount = float(input("Enter the amount for the item: "))
if itemAmount < 0:
continue
again = input("Do you want to add another item? Enter 'y' for yes and 'n' for no: ")
items.append(itemAmount)
if again == "y":
continue
elif again == "n":
numItems = len(items)
print(f"You purchased {numItems} items.")
sumAmount = sum(items)
print(f"The total for this purchase is {sumAmount} before tax.")
print(f"The average amount for this purchase is {sumAmount/numItems}.")
if numItems >= 10:
tax = (9/100)*sumAmount
else:
tax = (9.5/100)*sumAmount
print(f"You owe ${tax} in tax.")
break
else:
print("Invalid input")
continue
itemPrices()
while True:
user_input = input("type a number")
try:
if float(user_input) < 0:
print('this number is less than zero please try again')
continue
else:
print("good job this number is valid")
# place the code you want to execute when number is positive here
break
except ValueError:
print("this is not a number please enter a valid number")
continue
This question already has answers here:
Python input never equals an integer [duplicate]
(5 answers)
Closed 2 years ago.
I am trying to make a code guessing game where the user can choose the range of the code. The user tries to guess the randomly generated code until he/she gets it right. The computer also shows which digits the user gets correct. The problem is that when the user does guess the code correctly, the process just stops even though my codes says to print a congratulations message and go to the play again function. Please can anyone help? Thanks. Code:
import random
import string
def get_range():
Min = input("ENTER THE MINIMUM NUMBER THE CODE CAN BE: ")
Max = input("ENTER THE MAXIMUM NUMBER THE CODE CAN BE: ")
validate_range(Min, Max)
def validate_range(Min, Max):
Check_Min = Min.isdigit()
Check_Max = Max.isdigit()
if Check_Min is not True or Check_Max is not True:
print("INPUT MUST ONLY INCLUDE INTEGERS! ")
get_range()
elif Min == Max:
print("MINIMUM AND MAXIMUM NUMBER MUST NOT BE EQUIVALENT! ")
get_range()
elif Min > Max:
print("MINIMUM NUMBER MUST NOT BE GREATER THAN MAXIMUM NUMBER!")
get_range()
else:
Random = random.randrange(int(Min), int(Max))
get_guess(Random)
def get_guess(Random):
Guess = str(input("ENTER YOUR GUESS: "))
Check_Guess = Guess.isdigit()
if not Check_Guess:
print("INPUT MUST ONLY CONTAIN INTEGERS! ")
get_guess(Random)
else:
validate_guess(Guess, Random)
def validate_guess(Guess, Random):
Length = len(str(Random))
Digits_Correct = 0
if Guess == Random:
print("WELL DONE! YOU GUESSED THE NUMBER! ")
play_again()
else:
Digits = ["?"] * Length
for i in range(0, int(Length)):
if str(Guess)[i] == str(Random)[i]:
Digits[i] = Guess[i]
Digits_Correct += 1
else:
continue
if int(Length) > Digits_Correct > 0:
print("NOT QUITE! YOU GOT", Digits_Correct, " DIGITS CORRECT!")
print(Digits)
get_guess(Random)
elif Digits_Correct == 0:
print("NONE OF YOUR DIGITS MATCH! ")
get_guess(Random)
def play_again():
Choice = input("\n DO YOU WISH TO PLAY AGAIN? (Y/N)")
if Choice != "Y" or Choice != "N" or Choice != "y" or Choice != "n":
print("PLEASE ENTER A VALID INPUT! ")
play_again()
else:
get_range()
print("WELCOME TO CODE CRUNCHERS!\n ")
get_range()
I think the problem here is that your Guess is a string and your Random is an integer. To fix this, you can try to convert the Guess to an integer or the Random to a string.
Try this:
def validate_guess(Guess, Random):
Length = len(str(Random))
Digits_Correct = 0
if int(Guess) == Random:
print("WELL DONE! YOU GUESSED THE NUMBER! ")
play_again()
I think your problem is with types str and int. First of all your Min and Max are strings, so your line:
elif Min > Max: print("MINIMUM NUMBER MUST NOT BE GREATER THAN MAXIMUM NUMBER!")
does not work correctly. The other problem is that your variables Guess and Random are of different types, so Guess == Random will return False all the time.
Here's correct version of your code.
I've also added a few if cases to be able to quit the program without closing it.
import random
import string
def get_range():
Min = input("ENTER THE MINIMUM NUMBER THE CODE CAN BE: ")
Max = input("ENTER THE MAXIMUM NUMBER THE CODE CAN BE: ")
if Min == 'q':
return
validate_range(Min, Max)
def validate_range(Min, Max):
Check_Min = Min.isdigit()
Check_Max = Max.isdigit()
if Check_Min is not True or Check_Max is not True:
print("INPUT MUST ONLY INCLUDE INTEGERS! ")
get_range()
Min = int(Min)
Max = int(Max)
if Min == Max:
print("MINIMUM AND MAXIMUM NUMBER MUST NOT BE EQUIVALENT! ")
get_range()
elif Min > Max:
print("MINIMUM NUMBER MUST NOT BE GREATER THAN MAXIMUM NUMBER!")
get_range()
else:
Random = random.randrange(int(Min), int(Max))
get_guess(Random)
def get_guess(Random):
Guess = str(input("ENTER YOUR GUESS: "))
Check_Guess = Guess.isdigit()
if not Check_Guess:
print("INPUT MUST ONLY CONTAIN INTEGERS! ")
get_guess(Random)
else:
validate_guess(Guess, Random)
def validate_guess(Guess, Random):
Length = len(str(Random))
Digits_Correct = 0
if int(Guess) == Random:
print("WELL DONE! YOU GUESSED THE NUMBER! ")
play_again()
else:
Digits = ["?"] * Length
for i in range(0, int(Length)):
if str(Guess)[i] == str(Random)[i]:
Digits[i] = Guess[i]
Digits_Correct += 1
else:
continue
if int(Length) > Digits_Correct > 0:
print("NOT QUITE! YOU GOT", Digits_Correct, " DIGITS CORRECT!")
print(Digits)
get_guess(Random)
elif Digits_Correct == 0:
print("NONE OF YOUR DIGITS MATCH! ")
get_guess(Random)
def play_again():
print("\n DO YOU WISH TO PLAY AGAIN? (Y/N)")
Choice = input()
if Choice != "Y" or Choice != "N" or Choice != "y" or Choice != "n":
print("PLEASE ENTER A VALID INPUT! ")
play_again()
else:
get_range()
print("WELCOME TO CODE CRUNCHERS!\n ")
get_range()
This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 2 years ago.
My goal is creating a factorial program in python that asks infinite user input to find factorial of the number user type, until I would like to quit the program. But there is probably discrepancy between the lines of the code to works for exit the program and integer numbers below it.
1) I tried to solve this to not write int(input) I wrote just
input('Enter a number. Type exit to stop:> ')
both above or below the while True statement but it didn't work.
2) I also want to use lower() function to quit the program but when I use it, the discrepancy happens again because I ask user input for an integer but when I turn it to a normal input and type it the above while True statement, problem occurs.
3) Also I want to user input as a number with using that isdigit() function tried to use like this but it didn't work well:
factorial = 1
user_input = input('Enter a number: ')
while user_input.lower() != 'exit':
while not user_input.isdigit():
print('This is not a number. Please try again:> ')
user_input = int(input('Try again:> '))
user_input = int(input('Enter a new number: '))
.
.
.
and this, too didn't work
My actual code is this:
while True:
factorial = 1
user_input = int(input('Enter a number. Type exit to stop:> '))
if user_input == 'exit':
print('You are quitting the program')
break
elif user_input < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_input == 0:
print("The factorial of 0 is 1")
else:
for i in range(1, user_input + 1):
factorial = factorial * i
print("The factorial of", user_input, "is", factorial)
The program works like this:
Enter a number. Type exit to stop:> 4
The factorial of 4 is 24
Enter a number. Type exit to stop:> 5
The factorial of 5 is 120
Enter a number. Type exit to stop:> 6
The factorial of 6 is 720
and when I type 'exit' to quit from program I am receiving this kind of error:
Traceback (most recent call last):
File "E:/Kodlar/Python/Taslak projeler/taslak177.py", line 5, in <module>
user_input = int(input('Enter a number. Type exit to stop:> '))
ValueError: invalid literal for int() with base 10: 'exit'
As you can see, code blocks work instead of quitting the program with user input. How can I fix this?
Can anyone help? Thanks already!
Edit: I reorganized the code and it works perfectly fine. Thanks for your all responses!
while True:
user_input = input("Enter a number:> ")
if user_input == "exit":
print('You are quitting the program...')
break
else:
try:
factorial = 1
user_input = int(user_input)
if user_input < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_input == 0:
print("The factorial of 0 is 1")
else:
for i in range(1, user_input + 1):
factorial = factorial * i
print(f'The factorial of {user_input} is {factorial}')
except ValueError:
print("Please provide a valid number")
You should check if the input is exit before converting it to int and if so, break the loop.
Try this instead:
while True:
user_input = input("Enter a number:")
if user_input == "exit":
print('You are quitting the program')
break
else:
try:
user_number = int(user_input)
if user_number < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_number == 0:
print("The factorial of 0 is 1")
else:
# calc factorial here
except ValueError:
print("Please provide a valid number")
Your program get int inputs,
user_input = int(input('Enter a new number: '))
try this instead, get string input
user_input = input('Enter a new number: ')
and convert it into int later
user_input = int(user_input)
Because you are casting the user's response (a string) into an int in both cases.
user_input = int(input('Enter a new number: '))
and later
user_input = int(input('Enter a number. Type exit to stop:> '))
Perhaps try a little tweak:
while True:
factorial = 1
user_input = input('Enter a number. Type exit to stop:> ')
if user_input.lower().strip() == 'exit':
print('You are quitting the program')
break
elif user_input.isnumeric() and user_input < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_input.isnumeric() and user_input == 0:
print("The factorial of 0 is 1")
elif user_input.isnumeric():
for i in range(1, user_input + 1):
factorial = factorial * i
print("The factorial of", user_input, "is", factorial)
else:
print("Please enter an integer or 'exit'")
You could also wrap another if so you don't duplicate the isnumeric() tests
If the user gets to the end of the program I want them to be prompted with a question asking if they wants to try again. If they answer yes I want to rerun the program.
import random
print("The purpose of this exercise is to enter a number of coin values")
print("that add up to a displayed target value.\n")
print("Enter coins values as 1-penny, 5-nickel, 10-dime,and 25-quarter.")
print("Hit return after the last entered coin value.")
print("--------------------")
total = 0
final_coin = random.randint(1, 99)
print("Enter coins that add up to", final_coin, "cents, on per line")
user_input = int(input("Enter first coin: "))
total = total + user_input
if user_input != 1 and user_input!=5 and user_input!=10 and user_input!=25:
print("invalid input")
while total != final_coin:
user_input = int(input("Enter next coin: "))
total = total + user_input
if total > final_coin:
print("Sorry - total amount exceeds", (final_coin))
if total < final_coin:
print("Sorry - you only entered",(total))
if total== final_coin:
print("correct")
You can enclose your entire program in another while loop that asks the user if they want to try again.
while True:
# your entire program goes here
try_again = int(input("Press 1 to try again, 0 to exit. "))
if try_again == 0:
break # break out of the outer while loop
This is an incremental improvement on the accepted answer:
Used as is, any invalid input from the user (such as an empty str, or the letter "g" or some such) will cause an exception at the point where the int() function is called.
A simple solution to such a problem is to use a try/except- try to perform a task/ code and if it works- great, but otherwise (except here is like an else:) do this other thing.
Of the three approaches one might try, I think the first one below is the easiest and will not crash your program.
Option 1: Just use the string value entered with one option to go again
while True:
# your entire program goes here
try_again = input("Press 1 to try again, any other key to exit. ")
if try_again != "1":
break # break out of the outer while loop
Option 2: if using int(), safeguard against bad user input
while True:
# your entire program goes here
try_again = input("Press 1 to try again, 0 to exit. ")
try:
try_again = int(try_again) # non-numeric input from user could otherwise crash at this point
if try_again == 0:
break # break out of this while loop
except:
print("Non number entered")
Option 3: Loop until the user enters one of two valid options
while True:
# your entire program goes here
try_again = ""
# Loop until users opts to go again or quit
while (try_again != "1") or (try_again != "0"):
try_again = input("Press 1 to try again, 0 to exit. ")
if try_again in ["1", "0"]:
continue # a valid entry found
else:
print("Invalid input- Press 1 to try again, 0 to exit.")
# at this point, try_again must be "0" or "1"
if try_again == "0":
break