CODEWITHSUNDEEP
pythonpython-3.xauthentication
coconutfury
coconutfury

Reputation: 15

I'm making my first Python program using Python 3 but for some reason it gives me an error

I wanted to test my program in the middle of writing it, but it won't start for some reason

CODE:

def main():
    menu()
def menu():
    print("1.Login\n2.Register\n\n1.Help")
    input = int(input("Enter the corresponding number to the action you would like executed: "))
    if input==3:
        help()
    if input==2:
        register()
    if input==1:
        login()

def help():
    print("\nIf you enter the number ONE (1) You will be prompted to login to the system\nIf you enter the number TWO (2) you will be prompted to register, so you can login")

def register():
    usernameR = input("Enter your username: ")
    with open("usernamesR.txt","wt") as output:
        output.write(usernameR)
    print("Your username is: "+usernameR)
    passwordR = str(input("\nEnter your password: "))
    with open("passwordsR.txt","wt") as output:
        output.write(passwordR)
    print("Your password is: "+passwordR)

def login():
    print("")

if __name__=="__main__":
    main()

This is the error I got:

1.Login

Traceback (most recent call last):

2.Register

File "C:/Users/Joseph/PycharmProjects/LearningPython/Login_System.py", line 30, in main()

1.Help

File "C:/Users/Joseph/PycharmProjects/LearningPython/Login_System.py", line 2, in main

menu()

File "C:/Users/Joseph/PycharmProjects/LearningPython/Login_System.py", line 5, in menu input = int(input("Enter the corresponding number to the action you would like executed: "))

UnboundLocalError: local variable 'input' referenced before assignment

Upvotes: 1

Views: 59

Answers (2)

Vivek Sable
Vivek Sable

Reputation: 10223

input() ins in build function,

In code you give variable name as input, give different variable name.

def menu():
    print("1.Login\n2.Register\n\n1.Help")
    input = int(input("Enter the corresponding number to the action you would like executed: "))
    # ^^^^

Upvotes: 0

Tom Dalton
Tom Dalton

Reputation: 6190

Your variable name input clashes with the built-in method input. The compiler is getting confused, thinking that the input() call is calling your locally-bound variable, and then complaining that you haven't defined it yet. Changing your variable name to user_input should fix this.

Upvotes: 0

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