I'm working on a password checker that checks if the string is a valid password. I have to check if there is at least eight characters, must consist of only letters and digits and the last two characters must be digits.
It all seems to work so far other than the password.isdigit(). sometimes the password comes out valid and sometimes it doesn't. Any suggestions?
# Gets the users password
password = input('Enter a string for password: ')
# Splices the last two characters of the password
lastTwo = password[-2:]
# Checks the password if it is less than 8 characters
while len(password) < 8:
print('The password you entered is too short.')
print()
password = input('Enter a string for password: ')
# Checks the password if it is composed of letters and numbers
while password.isalnum() == False:
print('Your password has special characters not allowed.')
print()
password = input('Enter a string for password: ')
# Checks the spice to verify they are digits
while lastTwo.isdigit() == False:
print('Your last two characters of your password must be digits.')
print()
password = input('Enter a string for password: ')
print('Your password is valid.')
There are a handful of issues with your provided code. Particularly, you only check the subsequent rules while len(password) < 8. If you give it a password of length 10, the rules are never checked. Additionally, you don't update the lastTwo with each new password attempted
One way to fix this would be to replace your several while statements with if...elif..elif...else... wrapped in an overall while statement, as follows:
# Gets the users password
password = input('Enter a string for password: ')
while True:
# Checks the password if it is less than 8 characters
if len(password) < 8:
print('The password you entered is too short.')
# Checks the password if it is composed of letters and numbers
elif not password.isalnum():
print('Your password has special characters not allowed.')
# Checks the spice to verify they are digits
elif not password[:-2].isdigit():
print('Your last two characters of your password must be digits.')
else:
# we only get here when all rules are True
break
print()
password = input('Enter a string for password: ')
print('Your password is valid.')
This should work as you intended it. But while we're at it, why not tell the user every rule their password has broken? From a UI point of view, it helps to keep the user informed.
If we store an information message alongside whether the relevant rule has been met, we can quickly work out all of the rules that have been broken, like so:
valid_password = False
while not valid_password:
# Get a password
password = input('\nEnter a string for password: ')
# applies all checks
checks = {
'- end in two digits': password[-2].isdigit(),
'- not contain any special characters': password.isalnum(),
'- be over 8 characters long': len(password) > 8
}
# if all values in the dictionary are true, the password is valid.
if all(checks.values()):
valid_password = True
# otherwise, return the rules violated
else:
print('This password is not valid. Passwords must:\n{}'.format(
'\n'.join([k for k, v in checks.items() if not v])))
print('Your password is valid.')
You never update your value of lastTwo inside your while loop. Thus imagine if a user first entered a password abc123. Then lastTwo would be calculated as 23.
Now your code would find that the password is too short and prompt the user for a new password. Suppose he enters abcdefgh. This now passes your first and second checks. Notice however that lastTwo is still 23, and thus your third check will incorrectly pass.
You should thus recalculate the value of lastTwo whenever you accept a new password or directly check like this:
while (password[-2:]).isdigit() == False:
Related
I make password genrator using python then the teacher told to create a code that make the user choose what he want in the password
the password contains
1- lower case characters
2- upper case characters
3- numbers
4- punticuation marks
the teacher want me to make the user choose if he want punticution or not
if he dont want it have to be deleted from the password
i tried so hard but i got stuck
from msilib import change_sequence
import string
import random # I import this line to make the password unorgnaized
# this code below is all the characters we need for the password
s1 = list(string.ascii_lowercase)
s2 = list(string.ascii_uppercase)
s3 = list(string.digits)
s4 = list(string.punctuation)
# this code is for the user to put how much characters he need
characters_number = input("how many characters for the password?:")
# this while loop code is to make the user choose only 6 characters and up
while True:
try:
characters_number = int(characters_number)
if characters_number < 6 :
print("you need at least 6 characters") # if the user choose an letter or digits it will not allow him
characters_number = input("please enter the number again:")
else:
break # I break the loop here if the user write the correct data
except: #this code here if the user enter anything except numbers
print("please enter numbers only")
characters_number = input("how many characters for the password?:")
# the random and shuffle is to make the password unorgnized
random.shuffle(s1)
random.shuffle(s2)
random.shuffle(s3)
random.shuffle(s4)
# the password that will appear to the user it contains upper and lower letters, numbers and digit
# part1 here is for the letters I allocated 30% of the password for letters
# part2 here is for the digits and numbers I allocated 20% of the password for letters
part1 = round(characters_number * (30/100))
part2 = round(characters_number * (20/100))
password = []
for x in range(part1):
password.append(s1[x])
password.append(s2[x])
#the for loops here here is to create the password
for x in range(part2):
password.append(s3[x])
password.append(s4[x])
#this code here is to transform the password for list method to a string
password = "".join(password[0:])
print(password)
You can write a function like this:
def isUserNeedPunctuationMarks():
result = None
while True:
inp = input("Do you need punctuation marks in your password? (y/n)")
inp = inp.lower()
if inp == "y":
result = True
break
elif inp == "n":
result == False
break
else:
print("You must enter y or n")
return result
The above function will return True if the user needs punctuation marks, and False otherwise. User has to type y (for yes) or n(for no).
You can call isUserNeedPunctuationMarks() in the right place in your code. For example, removing punctuation marks from password string, you have to write a condition like this:
import re
if not isUserNeedPunctuationMarks():
password = re.sub(r'[^\w\s]', '', password) # This removes punctuation marks from a string using regex
Hope this helps :)
I'm trying to write a Python program that prompts the user for a password. It must meet the following requirements:
no less than 6 characters in length
no more than 12 characters in length
at least 1 numerical digit
at least 1 alphabetical character
no spaces
I can get through requirements 1-3, but as soon as I put in requirement 4, it stops working. I haven't even gotten to requirement 5 because I'm currently stuck. Any help is greatly appreciated! TIA!
Here is my code:
# --- Main ----------
def main():
#display student info
studentInfo()
#display welcome message
welcomeMsg()
#prompt user for a password
passWord = input("\nPlease create a password:\n")
#call function for password length validation
passWord = correctPW(passWord)
# --- Functions ----------
#student info
def studentInfo():
print("\nName:\tNAME")
print("Class:\tCMIS102")
print("Date:\t26 July 2022")
#welcome message
def welcomeMsg():
print("\nThis program will prompt the user to enter a password with the following requirements:")
print("\t- No less than 6 characters in length")
print("\t- No more than 12 characters in length")
print("\t- No spaces")
print("\t- At least one numerical digit")
print("\t- At least one alphabetical character")
#validate password requirements
def correctPW(passWord):
#check for minimum character requirement
while (len(passWord) < 6) or (len(passWord) > 12):
print("\nSorry! Your password is invalid.")
print("It must be no less than 6 characters and no more than 12 characters in length.")
passWord = input("\nPlease create a password:\n")
#check for one numerical digit and alphabetical character requirement
while (passWord.isdigit() < 1):
print("\nSorry! Your password is invalid.")
print("It must contain at least one numerical digit.")
passWord = input("\nPlease create a password:\n")
while (passWord.isalpha() < 1):
print("\nSorry! Your password is invalid.")
print("It must contain at least one alphabetical character.")
passWord = input("\nPlease create a password:\n")
#display if all password requirements are met
if (len(passWord) >= 6) and (len(passWord) <= 12) and (passWord.isdigit() >= 1) and (passWord.isalpha() >= 1):
print("\nCongratulations! Your password is valid!")
# --- Execute ----------
main()
The isdigit function will return True if all of the char in the password be numbers, like '65' or '1235' but if in the password, the user has any char near the numbers, like 'asdf3567' or '1a2b3c4d' or etc, it will return False, so this condition password.isdigit() < 1 is not a good condition
The isalpha function will return True if all of the char in the password be alphabets, like 'abc' or 'simplepassword' but if in the password, user has any char near the alphabets, like 'asdf3567' or '1a2/$b3c4d' or etc, it will return False, so this condition password.isalpha() < 1 is not a good condition
str.isdigit(): Return True if all characters in the string are digits and there is at least one character, False otherwise.
—https://docs.python.org/3/library/stdtypes.html#str.isdigit
Therefore, if passWord is (for example) "a1", then passWord.isdigit() will return False. You can, however, use it for each characters in the string, by using list (or generator) comprehension.
def check(password):
if not (6 <= len(password) <= 12):
print("Length must be between 6 and 12 (inclusive).")
return False
if any(char.isdigit() for char in password):
print("Must have a digit.")
return False
if any(char.isalpha() for char in password):
print("Must have an alphabet.")
return False
if any(char.isspace() for char in password):
print("Must not have a space.")
return False
return True
while True:
password = input("Enter a password: ")
if check(password):
break
print(f"Your password is: {password}... Oops, I said it out loud!")
If you want the actual number of digits in the password, you can use sum instead of any:
if sum(char.isdigit() for char in password) < 1:
Based on help(str) in python shell:
isdigit(self, /)
Return True if the string is a digit string, False otherwise.
A string is a digit string if all characters in the string are digits and there
is at least one character in the string.
isalpha(self, /)
Return True if the string is an alphabetic string, False otherwise.
A string is alphabetic if all characters in the string are alphabetic and there
is at least one character in the string.
So in lines:
while (passWord.isdigit() < 1): ...
while (passWord.isalpha() < 1): ...
passWord.isdigit() and passWord/.isalpha() even for your valid inputs are always False and for both of them < 1 will be True
and this will cause an infinite loop in your program.
I applied two new functions to your code to check if user inputs a valid pass you expected or not.
# --- Main ----------
def main():
#display student info
studentInfo()
#display welcome message
welcomeMsg()
#prompt user for a password
passWord = input("\nPlease create a password:\n")
#call function for password length validation
passWord = correctPW(passWord)
# --- Functions ----------
#student info
def studentInfo():
print("\nName:\tJessica Graeber")
print("Class:\tCMIS102")
print("Date:\t26 July 2022")
#welcome message
def welcomeMsg():
print("\nThis program will prompt the user to enter a password with the following requirements:")
print("\t- No less than 6 characters in length")
print("\t- No more than 12 characters in length")
print("\t- No spaces")
print("\t- At least one numerical digit")
print("\t- At least one alphabetical character")
def containsLetterAndNumber(input_password):
return input_password.isalnum() and not input_password.isalpha() and not input_password.isdigit()
def containsBlankSpace(input_password):
return (' ' in input_password)
#validate password requirements
def correctPW(passWord):
#check for minimum character requirement
while (len(passWord) < 6) or (len(passWord) > 12):
print("\nSorry! Your password is invalid.")
print("It must be no less than 6 characters and no more than 12 characters in length.")
passWord = input("\nPlease create a password:\n")
#check for one numerical digit and alphabetical character requirement
while not containsLetterAndNumber(passWord):
print("\nSorry! Your password is invalid.")
print("It must contain at least one alphabetical character and one numerical digit.")
passWord = input("\nPlease create a password:\n")
while containsBlankSpace(passWord):
print("\nSorry! Your password is invalid.")
print("It shouldn't have any blank space in it.")
passWord = input("\nPlease create a password:\n")
# display if all password requirements are met
if (len(passWord) >= 6) and (len(passWord) <= 12) and containsLetterAndNumber(passWord) and (not containsBlankSpace(passWord)):
print("\nCongratulations! Your password is valid!")
# --- Execute ----------
main()
Write a program that asks the user to enter a password (with the prompt "Password:"). It should then reward the user with "Congratulations, you're in!" if the password is between 8 and 20 characters long (inclusive), ends with a digit, and contains a period or comma. Otherwise, it should say "Wrong! This incident will be reported!"
I need to be able to have a number at the end and contain either a period or a comma. I am not sure on how to isolate these two parts of the password. What do I do to have the user asked to enter a password?
user_pass = str(input())
if (len(user_pass <= 8) and (len(user_pass >= 20)) and
print ("Congratulations, you're in!")
else:
print ('Wrong! This incident will be reported!')
Just add more conditions using and.
Python 3:
password = input("Password: ")
if (8 <= len(password) <= 20) and password[-1].isdecimal() and any(x in password for x in {'.', ','}):
print("Congratulations, you're in!")
else:
print("Wrong! This incident will be reported!")
Python 2:
password = raw_input("Password: ")
if (8 <= len(password) <= 20) and unicode(password[-1]).isdecimal() and any(x in password for x in {'.', ','}):
print("Congratulations, you're in!")
else:
print("Wrong! This incident will be reported!")
A hint:
def password_is_good(password):
return (
password_length_is_correct(password) and
password_ends_with_digit(password) and
password_contains_punctuation(password)
)
def password_length_is_correct(password):
# implement this, and the rest.
password = str(input())
if password_is_good(password):
# do something.
else:
# do something else.
You can access string elements by index, with negative indexes counting from the end; e.g. password[-1] would access the last character.
You can use the in operator to check for a character to be present in a string, like if 'c' in 'abcdef'.
For my homework I need to make a user registration system in python. I need to make a password section.
What I am meant to do?.
Here is my code, I can't work out whats wrong:
password = input("Please enter your password: ")
passwordl = len(password)
while passwordl <= 8:
print("password must be more than 8 characters - please try agian")
password = input("Please enter your password: ")
passworda = password.isalpha()
passwordi = password.isdigit()
while passworda != True or passwordi != True:
print("password needs to contain digits and characters - please re-enter")
password = input("Please enter your password: ")
The code is in a function btw.
Thanks
The functions isalpha() and isdigit() return true if all the members in the string are alphanumeric and digits, respectively.
What you need to do instead is to check if any characters in the string have the right properties, for example like this:
passworda = any([x.isalpha() for x in password])
passwordi = any([x.isdigit() for x in password])
Additionally, you need to redo all the checks (both length and character set checks) each time when the password is entered anew.
Instead of following
Password = input("Please enter your password")
In the last while loop with the checks over again you would be better just calling the password function as it's more efficient.
I'm receiving an error in python that says I, "Can't assign to literal". I was hoping that somebody could help me understand why this is coming up. Here's my code:
# Chapter 3 Challenge part 3
# use the code below and the following
# directions to see if you can solve the problem
# Your Co-Worker has been asked to write a
# 1st time "Log-in" program in which a user has to
# sign in with their user name. If they enter
# the correct id name they need to be asked
# to create their password.
# They were required to make sure the user did
# not leave the username entry blank and must have the
# correct user name or the progam should re-ask
# them to sign in... They cannot figure out what
# to do, so they have called on you to help...
# Here is what they have so far...
# They started their program like this....
# and included some of their documentation.
# Get username, check to see if not empty,
# then verify that username is correct.
# using username "jalesch" as test user.
name = False
while not name:
name = raw_input("Enter your user name: ")
print "Welcome," , name , "you must create a password."
password = False
while not password:
password = raw_input("""Enter a password.
Password must be at least 6 characters in length,
and must include one of these symbols (!, $, ?, &): """)
while len(password) < 6:
for "!" in password or "$" in password or "?" in password or "&" in password:
password = raw_input("""Enter a password.
Password must be at least 6 characters in length,
and must include one of these symbols (!, $, ?, &): """)
print "This has met the requirements."
check = raw_input("Please re-enter your password:")
while check != password:
print "You have entered the wrong password"
check = raw_input("Please re-enter your password:")
print "Welcome to the program!"
# prompt user to create a password that meets the
# following requirements:
# - Must be at least 6 characters in length
# - Must include at least one of these 4 symbols
# - ! $ ? &
# Once the user has entered their password, the program
# must check to see if it meets the requirements.
# If the requirements are not met, it should reask them
# to put in a password that meets the requirements
# until they get it correct.
# Test password is: $program!
# If it meets the requirements then prompt the user
# to re enter the password, and check that the
# first and second entry matches.
# If the 1st and 2nd match. display "Welcome to the program"
# If the 1st and 2nd don't match, re-prompt for the password
# until they get a correct match.
I also believe that the code is malfunctioning when I try to find the special characters, because it is skipping one of the while loops. Any help is appreciated, thanks.
on this line:
for "!" in password or "$" in password or "?" in password or "&" in password:
You probably meant to use if:
if "!" in password or "$" in password or "?" in password or "&" in password:
for x in y iterates over y, assigning each member to x sequentially, running the following indented code in a loop for each member of y. You can't assign a value to "!" because that's a literal character, not a variable.
def get_name(prompt):
while True:
name = raw_input(prompt).strip()
if name:
return name
def get_password(prompt):
while True:
name = raw_input(prompt).strip()
if len(name) >= 6 and (set(name) & set("!$?&")):
return name
name = get_name("Enter your user name: ")
print("Welcome, {}, you must create a password.".format(name))
while True:
print("Now enter a password. It must be at least 6 characters in length, and must include a symbols (!, $, ?, &).")
pass1 = get_password("Enter a password: ")
pass2 = get_password("Please re-enter your password: ")
if pass1 == pass2:
print("Welcome to the program")
break
else:
print("No match! Please try again")