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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
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When I run this easy program and want to exit it, I need to type exit 2 times (at line 7).
def registration():
def username_registration():
entering_username = True
while entering_username:
print("Type exit to leave the registration.")
usernam_from_registration = input("Enter username: ")
if usernam_from_registration == "exit":
break
lenght_usernam_from_registration = len(usernam_from_registration)
if lenght_usernam_from_registration > 15:
print("too long")
else:
return usernam_from_registration
username_registration()
print(username_registration())
registration()
Why is this and how can I make it so I only need to write it one time?
This is because you are calling the username_registration() function twice.
username_registration()
print(username_registration())
The first time you call it, nothing happens because you are not doing anything with the result.
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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.
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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))
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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.
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Closed 9 years ago.
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Python Question: i need to run a program that asks for a password but if the wrong answer is input three times the user is thrown out of the program i can run it in a while loop but cant get it to quit if the wrong password is entered.
Thanks for your help
Adding an approximation of how I'd do it, in the absence of an example containing the problem. else on a for loop will only execute if you did not break out of the loop. Since you know the max number of times to run the loop is 3 you can just use a for loop instead of a while loop. break will still break you out early.
for _ in range(3):
if raw_input("Password:") == valid_passwd: # really should compare hashed values (as I shouldnt have passwords stored in the clear
print "you guessed correctly"
break
print "you guessed poorly"
else:
print "you have failed too many times, goodbye"
sys.exit(1)
# continue on your merry (they got the right password)
How about sys.exit()
>>> import sys
>>> guess = False
>>> if guess:
... pass
... else:
... sys.exit()
http://docs.python.org/3/library/sys.html