I am writing a small login script at school, but I'm having trouble with a certain line of code. Here is the full thing:
I have asked my teacher but she isn't quite sure herself.
#!/bin/python3
def login ():
username = input ('username')
password = input ('password')
if username == 'TestAcc'*
if password == 'spectretest':
print ('Welcome to the SpectreOS developer test system')
else print ('invalid password')
else print ('invalid username')
*I get an error message on this line and I am not sure of the problem. Thanks for your help. :)
Your code has several syntax issues, use the following:
def login ():
username = input ('username')
password = input ('password')
if username == 'TestAcc':
if password == 'spectretest':
print ('Welcome to the SpectreOS developer test system')
else: print ('invalid password')
else:
print ('invalid username')
Keep in mind that indentation in Python matters a lot. I strongly suggest you check more about Python's syntax. You can check many tutorials (articles, videos) out there in the web.
Related
Just started learning python a couple days ago and have been trying to use what code I know to practice a basic code of asking for a user name and password from a list. I know there are far better/cleaner/matching user to password inputs but I'm just playing with what I know at this point.
users = ['Jon','Joe', 'Jole']
user_input = input('Username: ')
while user_input != users:
user_redo = input("I'm sorry but we dont recognize you. Please try another username: ")
this is where my problem is. Is there a simple way of breaking the loop if the user enters a matching username from the list?
passwords = ['donkey808','Hanna5006']
password = input('Password: ')
I guess the same question would apply to the password entry as well
while password != passwords:
pw_redo = input(f'Please enter correct password for user {user_input}: ')
else:
print(f'Access Granted {user_input}')
Write it like this.
users = ["Jon","Joe", "Jole"]
while 0 < 1 :
user_input = input('Username: ')
if user_input not in users:
print("I'm sorry but we dont recognize you. Please try another username: ")
elif user_input in users:
break
Just do while user_input not in users:
not in checks if user_input is literally not in users
It might also be better to do if user_input not in users:, I don't see the point of a while.
I've created a socket with a login system. The code below is my login class on my client system. it receives the message login successful or username/password incorrect. the print statement shows that this is getting through fine from the server so my issue isn't there or on the server part. when a password is wrong it starts back to login which is what I want it to do.the issue is when it's a correct login. the server will send login successfully and that will print but it won't break out of the loop and will just continue to ask for my username and password.
def login():
username = str(input("Please enter your username to login: "))
password = str(input("Please enter your password to login: "))
joint =(username+password)
message = hashPW(joint)
s.send(message.encode())
ls = s.recv(1024)
print(ls)
while ls != "Login successfull":
login()
else:
after_choice()
server code: I don't think this is where the problem lies. I'm pretty sure it's with the client and me being unable to break it out of the loop properly or something but I figured id add it anyway to help.
def recive():
password = c.recv(1024).decode()#reciving the password
print("Password from client: ", password)
PWcheck()
def PWcheck():
f= open("passwords.txt")
f1=f.readlines()
if password in f1:
print("Login successfull")
ls =("Login successfull")
c.send(ls.encode())
wait() #just the next class
else:
print("Username or password incorrect")
ls =("Username or password incorrect")
c.send(ls.encode())
recive()
if anyone needs more code adding i can. any help would be greatly appreciated.
sorry for my nooby question; I had a python application with only functions but now I have to revise it with classes that I am not expert on.
Here is my basic user Class;
class User:
def __init__(self,username,password,basket):
self.username=username
self.password=password
self.basket=basket
and the input function that I want to put inside the class;
def Login(self):
while True:
username = raw_input('\nPlease log in by providing your user credentials \nUser Name :')
if (username == users[0]['username'] or username == users[1]['username']):
password = raw_input('Password : ')
if (password == users[0]['password'] or password == users[1]['password']):
print('Successfully logged in!')
print (
'Welcome,' + username + '! Please choose of the following options by entering the corresponding menu number.')
global LoggedUserName
LoggedUserName = username;
return True;
break;
else:
print('Your password is not correct. Please try again!')
else:
print ('Your username is not correct. Please try again!')
Thank you for your precious helps!
You need to make instance of your class like that
Aname=User()
Aname.login()
But im not sure im too newbie object orriented programming
I am trying to create a simple login system. What I'm doing is storing the login data in a text file called 'accounts.txt'
Now, when user tires to login, it first checks if the username given by the user is in the 'accounts.txt'. If it exists, then it asks for the password and then checks if password matches with the password in 'accounts.txt'
fr = open('accounts.txt', 'r')
while True:
username = input('Enter your username: ') # Ask for their username
if username in fr.read(): # Check if username exists
password = input('Enter password: ') # Ask for password if username exists
if username+password in fr.read():
print('Welcome ' + username)
break
else:
print('Wrong password')
Note, the password save in accounts.txt is in the format of usernamepassword so if username is jack and password is gate, the actual password in the txt file will be jackgate, hence im using username+password to check if password is correct.
The problem occuring is if the user enters correct username, then program moves ahead properly but even if the password entered is right, it still displays 'Wrong password' .When the second time user enters username, it even shows error for wrong username. I tried to play with the code for a long time but couldn't come up with a solution. I guess it has something to do with fr.read(). Can I use that 'fr' object only once?
Let me suggest some improvements with my answer to your question. I would read the accounts file in its entirety so you have an in-memory structure. If you do this as a dictionary in the form accounts[USER] -> PASS you can easily check for any account as per the code below.
Regarding my suggestions (they do not exactly only answer your questions, but IMHO the topic of writing login code should be treated with care):
I strongly recommend not to store passwords in plain text, regardless of application importance, always use hashes.
Do not store just the password hash, always use salting.
Do not tell the person trying to log in, if the username or the password was wrong, always just say "that's not the right combination", thus making it harder to break in.
Please find information about hashing functions in Python here: https://docs.python.org/3/library/hashlib.html#randomized-hashing
This site has a good introduction on salting ans securing passwords: https://crackstation.net/hashing-security.htm
Do you users a favor and treat the username as no case-sensitive. That is a totally valid approach, but it annoys me every time I have to use such a site (just like email addr are not case-sensitive)
As I am a total layman regarding password security, maybe one of the other Stackoverflow users can jump in with a comment and expand on this topic.
Anyway, here is my answer for your question on how to check for a login. I created a function check_account() that returns True or False, depending on wether the supplied credentials were correct or not.
import hashlib
import os
import binascii
def check_account(usr, pwd):
# read the accounts file, a simple CSV where
# username, salt value and password hash are
# stored in three columns separated by a pipe char
accounts = {}
fr = open('/users/armin/temp/test.csv', 'r')
for line in [x.strip().split("|") for x in fr.readlines()]:
accounts[line[0].lower()] = (line[1], line[2])
fr.close()
# now go looking if we do have the user account
# in the dictionary
if usr in accounts:
credentials = accounts[usr]
# credentials is a list with salt at index 0
# and pwd hash at index 1
# generate the hash form the functions parameters
# and compare with our account
h = hashlib.blake2b(salt=binascii.unhexlify(credentials[0]))
h.update(pwd.encode('utf-8'))
if credentials[1] == h.hexdigest():
return True
else:
return False
else:
return False
def main():
while True:
username = input('Enter your username: ') # Ask for their username
password = input('Enter password: ') # Ask for password if username exists
if check_account(username.lower(), password):
print("Welcome, {0}".format(username))
else:
print('Username or password unknown')
if __name__ == '__main__':
main()
To create the data for a user account, use may this code.
def create():
username = input('Enter your username: ').lower() # Ask for their username
password = input('Enter password: ') # Ask for password if username exists
salt = binascii.hexlify(os.urandom(hashlib.blake2b.SALT_SIZE))
print("SALT value:", salt)
h = hashlib.blake2b(salt=binascii.unhexlify(salt))
h.update(password.encode('utf-8'))
print("Pwd hash:", h.hexdigest())
You can use startswith and endswith:
fr = [i.strip('\n') for i in open('accounts.txt')]
while True:
username = input()
if any(i.startswith(username) for i in fr):
password = input('Enter password: ')
if any(username+password == i for i in fr):
print("welcome")
break
else:
print("wrong password")
I would do
if password in fr.read():
instead of
if username+password in fr.read():
This is because for it to get to the if password in fr.read loop it first has to pass the if username in fr.read loop. However, the only problem I find with this is that if they enter a correct username but enter the wrong password for that username but correct password for another username it will still pass.
That is why I think you should use a dictionary not a text file.
For example, if the usernames allowed is username and username1 and the password is username and username1, then in a different .py file, you can say.
username_password={'username':'username','username1':'username1'}
that makes a dictionary that has the username and passwords.
let's say you name that file stuff.py. Then in the second file that has to be in the same directory, you can do
from stuff import * #imports all values from stuff.py
while True:
username = input('Enter your username: ') #gets username
if username_password.has_key(username):
password = input('Enter password: ')
if password== username_password[username]:
print('Welcome '+username)
break
else:
print('Wrong password')
break
else:
print('Wrong username')
I still don't get why you have a while loop, but if you want it, it is fine. Also, I added an else loop just in case the username is wrong.
I am trying to create a login.
I am not sure how to create/import a library of usernames and passwords; I am researching to find an answer at the moment but asked either way.
Match usernames with passwords (partially solved; need to add multiple usernames with matching passwords).
How to create a loop if password is incorrect? If incorrect password is entered the user needs to be prompted again to enter the password.
How to limit loop to certain number of attempts for password.
Below is what I have tried:
def check_password(user, password):
""" Return True if the user/pass combo is valid and False otherwise. """
# Code to lookup users and passwords goes here. Since the question
# was only about how to do a while loop, we have hardcoded usernames
# and passwords.
return user == "pi" and password == "123"
def login():
""" Prompt for username and password, repeatedly until it works.
Return True only if successful.
"""
try:
while True:
username = raw_input('username:')
password = raw_input('password:')
if check_password(username, password):
break
else:
print "Please try again"
print "Access granted"
return True
except:
return False
For testing: login().
This fixed lack of loop prompting if wrong password due to using return instead of print; and if instead of while.
def login():
#create login that knows all available user names and match to password ; if password is incorect returns try again and propmts for password again#
username = raw_input('username:')
if username !='pi':
#here is where I would need to import library of users and only accept those usernames; needs to be like 'pi' or 'bob' or 'tim'etc.
print'user not found'
username = raw_input('username')
password = raw_input('password:')
#how to match password with user? store in library ?
while password != '123':
print 'please try again' # You have to change the 'return' to 'print' here
password = raw_input('password')
return 'access granted'
#basically need to create loop saying 'try again' and prompting for password again; maybe smarter to ask limited number of
#times before returning 'you have reached limit of attempts#
if password == '123':
#again matching of passwords and users is required somehow
return 'access granted'
>>> login()
username:wronguser
user not found
usernamepi
password:wrongpass
please try again
password123
'access granted'
>>>
First attempt before updating thanks to Merigrim:
def login():
# Create login that knows all available user names and match to password;
# if password is incorect returns try again and propmts for password again#
username = raw_input('username:')
if username !='pi':
# Here is where I would need to import library of users and only
# accept those usernames; needs to be like 'pi' or 'bob' or 'tim'etc.
return 'user not found'
password = raw_input('password:')
# How to match password with user? store in library?
if password != '123':
return 'please try again'
password = raw_input('password:')
if password != '123':
return 'please try again'
# Basically need to create loop saying 'try again' and prompting
# for password again; maybe smarter to ask limited number of
# times before returning 'you have reached limit of attempts
elif password == '123':
# Again matching of passwords and users is required somehow
return 'access granted'
Here is how it currently works:
>>> login()
username:pi
password:123
'access granted'
>>> login()
username:pi
password:wrongpass
'please try again'
I need to create loop to prompt again for password.
What you want is the while statement.
Instead of nesting if-statements like this:
if password != '123':
return 'please try again'
password = raw_input('password:')
if password != '123':
return 'please try again'
elif password == '123':
return 'access granted'
You can do this:
while password != '123':
print 'please try again' # You have to change the 'return' to 'print' here
password = raw_input('password:')
return 'access granted'
This will continue prompting the user for a password until the right password is entered. If you want to become more familiar with the while statement, I suggest checking out some tutorials, like this one.
Please note that if you return something the function will exit there, so the user will never be prompted for a password. In the code above I changed the return to a print statement instead.
Here's another solution with the user name and password factored out, and an exception handler in case someone tries to abort the input.
Also, FYI it is best to take the user and password together so as not to let crackers know what is and is not a valid username.
def check_password(user, password):
""" Return True if the user/pass combo is valid and False otherwise. """
# Code to lookup users and passwords goes here. Since the question
# was only about how to do a while loop, we have hardcoded usernames
# and passwords.
return user == "pi" and password == "123"
def login():
""" Prompt for username and password, repeatedly until it works.
Return True only if successful.
"""
try:
while True:
username = raw_input('username:')
password = raw_input('password:')
if check_password(username, password):
break
else:
print "Please try again"
print "Access granted"
return True
except:
return False
# For testing
login()