Second iteration of While Loop creates an error [closed] - python

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I am very new to Python, I have searched and cannot find answer. Thank you in advance.
I purposely type the incorrect answer and it responds properly - prompts me to give input again.
I now type the correct answer and it responds with error (my code is below error)
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
in
54 print("Enter Y/N:")
55 answer = input()
---> 56 answer = input.upper()
57
58 print("Great! Lets get started.")
AttributeError: 'function' object has no attribute 'upper'
`# intro text
print("Are you ready to create a super hero with the super hero generator 3000?")
# Ask for input / use input() to take it / use upper() to adjust text
print("Enter Y/N:")
answer = input()
answer = answer.upper()
# adding while loop to condition input(answer)
while answer != "Y":
print("Sorry, but you have to choose Y to continue.")
print("Enter Y/N:")
answer = input()
answer = input.upper()
print("Great! Lets get started.") ```

Change your code to:
while answer != "Y":
print("Sorry, but you have to choose Y to continue.")
print("Enter Y/N:")
answer = input()
answer = answer.upper()
input() function has no function upper(), you want to call upper() on answer.

Related

Trying to find out how my code goes wrong [closed]

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I am a beginner learning Python and wrote the following codes. To control the loop flow I tried using a variable "test", but the loop never stopped. Appreciate if any thoughts.
test=False
while test==False:
a=input("Please enter an even number to make test True: ")
if int(a)%2==0:
test==True
print("test is now True")
else:
print("Please try again!")
test=False
while test is False:
a=input("Please enter an even number to make test True: ")
if int(a) %2 == 0:
test = True
print("test is now True")
else:
print("Please try again!")
Fixed the test == true, should've been test = true :) have fun learning
heres the right code:
test=False
while test==False:
a=input("Please enter an even number to make test True: ")
if int(a)%2==0:
test=True
print("test is now True")
else:
print("Please try again!")
here's what you did wrong:
in the 5th line test==True the == checks if that's what the variable true stores, it doesn't set it; so this can be simply fixed by test=True

Unable to print a sentence in python 3.7 [closed]

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I have created a function named user_choice() using Jupyter Notebook. This function is expecting an valid integer input from the user and prints the same. If a user any input other than integer then it should display an error message like " Sorry that is not a digit!" and it will again ask the user to enter valid input.
Below is my code for the function user_choice()
def user_choice():
choice = "WRONG"
while choice.isdigit() == False:
choice = input("Enter a digit(0-10): ")
if choice.isdigit == False:
print("Sorry that is not a digit!")
return(int(choice))
On calling the above function and entering non integer value, It is not displaying the message "Sorry that is not a digit!"
Enter a digit(0-10): ten
Enter a digit(0-10):
Looks like you just forget brackets after .isdigit method in if statement.
You need to use recursive call
def user_choice(choice="WRONG"):
if choice.isdigit() == False:
choice = raw_input("Enter a digit(0-10): ") #use input for higher python version, mine is 2.7 hence used raw_input
if choice.isdigit() == False:
print("Sorry that is not a digit!")
user_choice(choice)
else:
print("Captured:"+choice)
return(int(choice))
A shorter version:
def user_choice(choice="WRONG"):
if choice.isdigit() == False:
choice = raw_input("Input must be a digit(0-10): ") #use input for higher python version, mine is 2.7 hence used raw_input
user_choice(choice)
else:
print("Captured:"+choice)
return(int(choice))

Yes / No Function Always Returning False :( [closed]

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I'm completely new to programming and am trying to write a function that takes a yes or no question. However, when I run the below it always seems to complete with my variable as False. There has to be something simple here that I'm missing. I'd love to hear any thoughts / feedback / improvements to it. Thank you!
def yes_or_no(question):
answer = input(question).lower().strip()
print("")
while not(answer == "y" or answer == "yes" or \
answer == "n" or answer == "no"):
print("\nSorry, only Y or N please.")
answer = input(question).lower().strip()
print("")
print(answer)
if answer == 'y' or answer == 'yes':
answer = True
else:
answer = False
You forgot to return the answer, so that function returns None by default. Add return answer at the end of the function.

How can I define a Boolean to a variable? [closed]

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Closed 3 years ago.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
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I am trying to make a calculator. I have the code that does the operations working, but I am trying to allow the user to keep the calculator running without having to rerun the code. i an trying to assign a Boolean to variable, but python keeps telling me there is a name error, which is the the variable is not defined. Can you please help me?
Thank you in advance.
I have tried to change the available name but it doesn't do anything.
The code stopped working when it got to the while run == true line.
else:
print('Invalid operator, please run code again')
run = True
while run == True:
print(' do you need another problem solved? y/n')
if input() == y:
run = True
elif input() == n:
run = False
I expected the code to ask me if I need another problem solved, but there is a name error.
If the error is complaining about y or n it's because you need to surround it with quotes
if input() == "y":
run = True
Also, run == True is not needed
while run:
does the trick
else:
print('Invalid operator, please run code again')
run = True
while run:
print(' do you need another problem solved? y/n')
inp=input()
if inp == 'y':
run = True
elif inp == 'n':
run = False
you have used input 2 times so your code will wait 2 times for input, use input() once.

Weird python syntax error with variable [closed]

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Closed 5 years ago.
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I keep getting a syntax error on line 10 that says
Traceback (most recent call last):
File "main.py", line 5, in <module>
user = input("Choose a number")
File "<string>", line 1
python main.py
^
I can't figure it out. Can anyone help?
import random
number = random.randint(1, 20)
attempt = 5
def game():
global attempt
attempt -= 1
if attempt > 0:
if int(user) > number:
user = input("Choose a number")
print("Choose a smaller number\n " + (str(attempt) + " tries left"))
game()
elif int(user) < number:
print("Choose a larger number\n "+ (str(attempt) + " tries left"))
game()
elif int(user) == number:
print ("You guessed the number ")
else:
print("CONGRATULATIONS ")
game()
First, this is a run-time error, not a syntax error.
The problem is actually in the previous line: you've tried to use the value of user before it has one. You have to fix this before you can continue debugging your program. You need to get the user's first number before you can test it.
Problem A
I can't help but notice that your traceback message doesn't align with the code you have provided. It seems as though you already have the user variable assigned before defining and calling the game() function.
Solution A
If that is the case, try removing that line and apply the changes Prune recommended for you in the other answer. This will solve the code related problems in the program.
Problem B
However, the traceback also mentions that on line 1 you have python main.py. This is most likely caused by the way you are running the script.
Solution B
Try running it instead in the IDLE (distributed with Python) or Terminal / Command Prompt (depends on your OS) if not doing so already.
Last Resort
If the problem still arises, try running the snippet of code you included in the question by itself to see if it is functional ( add on Prune's revisions with it ).
Since the revised snippet should work, any errors that occur is definitely caused by how the code is ran. If no errors occur then look through your original source code for anything else that might be causing problems.
Good luck!

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