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How do I deal with unlimited loop problems?

so i have this code
pa=""
newpa=""
use=''
use=input('enter username:')
pa=input('enter password')
while True:
if len(pa)==12:
newpa=input('re-enter password:')
else:
print('the password used did not meet our system requirements, please enter a 12 digit password')
if newpa==pa:
print('you have successfully created a new account!')
whenever I insert a pa that's incorrect the else print keeps looping. How do I make it so whenever I enter a pa that doesn't meet the if to loop back to newpa
if this help this is my algorithm alogrythm of what I am trying to go for

First thing to do would be the put the input statement inside the while loop. At the moment your code loops back to the top of the while loop without giving the user a chance to re-enter their password.
You'll also want to move the check for the re-entry part so that it only executes if the first password was valid. And let the user know if their passwords didn't match.
Finally, you'll want to break the loop when a valid password is entered and verified. Use break for this.
The final code would look like this:
pa=""
newpa=""
use=''
use=input('enter username:')
while True:
pa=input('enter password')
if len(pa)==12:
newpa=input('re-enter password:')
if newpa==pa:
print('you have successfully created a new account!')
break
else:
print('the passwords did not match!')
else:
print('the password used did not meet our system requirements, please enter a 12 digit password')
Hope that helps and makes sense!

Your loop will run continuously until you add break statement.
You can try something like below,
pa=""
newpa=""
use=''
use=input('enter username:')
while True:
pa=input('enter password')
if len(pa)==12:
newpa=input('re-enter password:')
if newpa==pa:
print('you have successfully created a new account!')
break
else:
print('the password used did not meet our system requirements, please enter a 12 digit password')

When you enter in the last if you want to exit from loop. So, add break after print.
if newpa==pa:
print('you have successfully created a new account!')
break

Use a while True loop and break from that loop if the password meets the requirements. Then you can enter the password a second time and compare if both inputs match.
username = input('enter username:')
while True:
password = input('enter password')
if len(password) == 12 and password.isdigit():
break
print('the password used did not meet our system requirements, please enter a 12 digit password')
password_check = input('re-enter password:')
if password == password_check:
print('you have successfully created a new account!')
You might want to wrap that in an additional loop if the passwords don't match.
If you use functions you can reduce the complexity of the main part.
def enter_password(message):
while True:
password = input(message)
if len(password) == 12 and password.isdigit():
break
print('the password used did not meet our system requirements, please enter a 12 digit password')
return password
def main():
username = input('enter username:')
while True:
password = enter_password('enter password')
password_match = enter_password('re-enter password')
if password == password_match:
break
print("The passwords don't match.")
print('you have successfully created a new account!')
if __name__ == '__main__':
main()

Please use break to break the statement and continue to continue the loop statement according to your requirement. I hope the following code meets your requirement. Thanks and let me know if there's anything I can help with.
user=input('enter username:')
while True:
pa=input('enter password:')
if len(pa)==12:
newpa=input('re-enter password:')
while newpa!=pa:
newpa=input('re-enter password:')
else:
print('the password used did not meet our system requirements, please enter a 12 digit password')
continue;
if (newpa==pa):
print('you have successfully created a new account!')
break;

How do I Avoiding Looping back to lines in code?

I am self-teaching myself python and have run into a problem that I can not seem to find a way around.
I have created a piece of code that compares an entered password to one stored in a database.
My code should have two possibilities.
1) If the password is correct.
The user is prompted to enter a new password and then the prompt to enter the password must appear again (This time accepting the new password).
2)If the password is incorrect the user will be prompted to enter the password until the correct password is entered.
In VBS I used to be able to use the GOTO command.
I am not sure if this is available in Python and if it is I would like to avoid using it as it creates a very illogical hard to follow the program.
password = "#123"
entry = input("Please Input The Password:")
if (password == entry):
entry = input("Password correct you may enter a new password.")
else:
entry = input("Password Incorrect, Try again.")

There are various ways you could complete this. Here is a simple way you could achieve it using while loop and break statement.
password = "#123"
while(True):
entry = raw_input("Please Input The Password: ")
if (password == entry):
print("Password correct you may enter a new password.")
break
else:
print("Password Incorrect, Try again.")
Hope it helped.

while password != entry: # Executes until (password == entry), and does not execute if it is met, even for the first time.
print('Sorry, wrong password.')
entry = input('Enter password >') # or other source of data
print('Correct!')
Edit: additional ways you can do this:
while True: # forever loop, but
entry = input('Enter password >') # or other source of data
if password == entry:
print('Correct!') # you can also put this outside of the loop
break # exit the loop no matter what
# do not put code after the break statement, it will not run!
print('Sorry, wrong password') # will execute only if password != entry, break ignores the rest of the code in the loop

Easiest to make a function with a while statement.
password = "#123"
def login():
while True:
answer = input("Please Input The Password:")
if answer == password:
print("Password correct you may enter a new password.")
else:
print("Password Incorrect, Try again.")
break
login()

Python - Beginner Level - User input and while Loop combination

I have did a little research around "Google", "YouTube", "Facebook" and "Stack Overflow" and I haven't found what I was looking for. So, I need your guidance. :)
I want program to ask user input "PASSWORD" and every time user inputs wrong password the program asks password again, and again, and again until user
types the correct password. Let's say that the password is as simple as "abc123".
So, I start the program and it asks to input: "PASSWORD: ". If user types "abc123" then program prints "GOOD PASSWORD". If user types anything what is not "abc123" then program prints "BAD PASSWORD". As simple as that.. for now.
My best attempt:
#SECTION1_ASKING
passwordInput=input('PASSWORD: ')
password='abc123'
while password == passwordInput:
print('GOOD PASSWORD')
break
else:
print('BAD PASSWORD')
passwordInput=input('PASSWORD: ')
#SECTION2_RE-ASKING
while False:
while password == paswordInput:
print('GOOD PASSWORD')
else:
print('BAD PASSWORD')
passwordInput=input('PASSWORD: ')
but I either make password asked once or twice or I stuck in Infinite while Loop.

Try this:
passwordInput=raw_input('PASSWORD: ')
password='abc123'
while True:
if password == passwordInput:
print('GOOD PASSWORD')
break
else:
print('BAD PASSWORD')
passwordInput=raw_input('PASSWORD: ')

You can do as below in few lines.
password='abc123'
while(True):
passwordInput=input('PASSWORD: ')
if(passwordInput==password):
print("Good password")
break
else:
print("Bad password")

Here is my solution:
password = 'abc123'
while True:
passwordInput = input('PASSWORD: ')
if password == passwordInput:
print('GOOD PASSWORD')
break
else:
print('BAD PASSWORD')
How does that differ from yours? For a start you can see that I only have one call to input(), that's generally a good idea because then you only need to check the password in one place. Notice that I use if instead of while for the test.
In your second example you have while False:. Can this ever be True? No, so it won't get executed.
Notice as well that I use more whitespace. That does not slow the program down, it makes it easier to read. See also PEP008
Now you have it working, just for fun, consider an improvement. Normally you don't want the password to be visible when it is typed, but there's a module for that: getpass - it's part of the standard library so you don't need to download anything.
Add import getpass to the top of your script. Instead of input use getpass.getpass, in the same place in the program, with the same parameters. Now it won't echo the password entered by the user.
getpass is the module name, getpass is also a function name, so getpass.getpass('PASSWORD: ') means execute the getpass() function in the getpass module. We use lots of modules in Python, so its worth getting used to using them. You can find the documentation here, note there is also a getpass.getuser() to play with.

Making a login forum in Python 3 [closed]

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Closed 1 year ago.
Improve this question
Checker = 'Incorrect'
name = input('Please Enter Your Name: ')
password = input('Please Enter Your Password: ')
if password == 'badger123':
print('Checking Password...')
else:
print(Checker)
if Checker == 'Incorrect':
exit()
import time
time.sleep(5)
print('Password is Correct')
time.sleep(3)
input('Press Enter to Continue...')
print ('Hello' , name)
import time
time.sleep(1)
print ("Closing Python...")
import time
time.sleep(5)
input('Press Enter to continue...')
exit()
I need help , the correct password is badger123 but even if I enter the correct password it will print 'password is correct' and then exit but I want it to ignore exit and continue. When the wrong password is entered then the program will print incorrect and will exit. The programs reacts correctly when the incorrect password is entered but not when the correct password is entered. I NEED HELP!

The value of Checker never actually changes, so
if Checker == 'Incorrect':
exit()
always evaluates to True, which means exit() is always called.
One solution is to do the checking in the if statement
import time
password = input('Please Enter Your Password: ')
if password == 'badger123':
print('Password is Correct...')
else:
print('Password Incorrect')
exit()
time.sleep(3)
input('Press Enter to Continue...')
print ('Hello' , name)
time.sleep(1)
print ("Closing Python...")
input('Press Enter to continue...')

First you only need to import a module once (by convention at the top of your file), and not each time you want to use a function.
The problem you describe is impossible with the code you show
Checker = 'Incorrect'
if Checker == 'Incorrect':
exit()
Since you assign 'Incorrect' to Checker and never changes it after that, the rest of your code is not executed, and your program will never print 'password is correct', it will however print 'Checking Password...' when the correct password is entered and then exit because Checker is always the same, it's like doing if 1 == 1:exit()

Why it isn't working
You set value checker to "incorrect"
When the password is found to be correct, nothing is done about it. Nothing notifies the program in any way that the correct password has been entered. You can see in program that all the program does to acknowledge that the password is correct is simply say that it is "checking" and nothing else. 3 lines later the program will find that checker is still "incorrect" because the program, seeing that the password is correct does nothing to change checker so line 9 is inevitably going to happen.
Fix:
Put line 10 with the else statement and remove line 9. So your code will look something like.
password = input('Please enter your password: ')
if password == 'badger123':
print('Checking Password...')
else:
print(Checker)
exit()
I can suggest an improvement.
password = input('Please enter your password: ')
if not password == 'badger123':
print(Checker)
exit()
print('Checking Password...')

I think this would be the ANSWER to my question! , All I did is I changed the value of Checker. But now when the person gets the password wrong I want the program to automatically Restart . How do I do that?
Checker = 'Incorrect'
name = input('Please Enter Your Name: ')
password = input('Please Enter Your Password: ')
if password == 'badger123':
print('Checking Password...')
else:
print(Checker)
Checker = exit()
if Checker == 'Incorrect':
input(Checker)
import time
time.sleep(5)
print('Password is Correct')
time.sleep(3)
input('Press Enter to Continue...')
print ('Hello' , name)
import time
time.sleep(1)
print ("Closing Python...")
import time
time.sleep(5)
input('Press Enter to continue...')
exit()

Except error for a name

I have a problem with a program I am making to do with an arithmetic quiz.when I enter a name the quiz should print the welcome message however if I enter a blank or a number then the programme should loop the question but tell the user that they have made an error. I used the try statement and the NameError statement to try and resolve this problem but when I use the function "except" and run it in idle it says that I have an invalid syntax.
here is the code i am trying to fix:
import random
while True:
try:
UserName = input("What is your name:")
except NameError:
print ("The answer given does not compute!! Try again")
continue
else:
break
print(UserName," welcome to the Arithmetic Quiz.")
I have edited the code for the program, but still when I try to run it in Idle it highlights except and then says invalid syntax.

You don't need a try-except to check for an empty string
while True:
UserName = input("What is your name:")
if not UserName:
print("The answer given does not compute!! Try again")
continue
else:
break
print(UserName," welcome to the Arithmetic Quiz.")

This is what the proper indentation should be for the try: except: else: block
import random
while True:
try:
UserName = input("What is your name:")
except NameError:
print("The answer given does not compute!! Try again")
continue
else:
break
print(UserName," welcome to the Arithmetic Quiz.")
caveat (thanks cricket_007) this loop will never end. I had meant this comment to show how the try: indentation should have been for the OP.