I start to learn python a couple of days ago,Im trying to make some kind of login interface with multiple passwords options but can't figure out the right way to make it loop until I enter the right password ,I tried to use "while" but cant seem to figure out the syntax and placement with it in my code,I want it to repeat the first block of code if "else" is the result so I can try and type the password again,please help?
import random
male = random.choice([True, False])
import random
num = random.choice(["1", "0"])
name = "joe"
user_input= input ("insert Password here ")
if ((user_input == "joey") or (user_input == "loki")):
if male == True: print("hello")
if male == False: print("wellcome")
if name == "joe":
if num == "1":
print("hi world")
if name == "joe":
if num == "0":
print("nice")
if name + num == ("joe" + "0"):
print("thats working")
else:
print ("Wrong Password,Please try again.")
First of all I would like to point out that you only need to import your libraries once. Second if you want your code to repeat you will require a loop, in python there are 2 types of loops - for loop and while loop. For your requirement I would suggest using a while loop as given below. I couldn't really understand what you were trying to accomplish but I have improved your code a bit.
import random
male = random.choice([True, False])
num = random.choice(["1", "0"])
name = "joe"
while True:
user_input= input("insert Password here ")
if (user_input == "joey") or (user_input == "loki"):
if male == True:
print("hello")
elif male == False:
print("wellcome")
else:
break
if name == "joe" and num == "1":
print("hi world")
elif name == "joe" and num == "0":
print("nice")
if name + num == ("joe" + "0"):
print("thats working")
else:
break
else:
print ("Wrong Password,Please try again.")
Another method of having it run repeatedly is to use recursive functions, however since you are just starting out with python I would suggest learning loops before you get into recursive functions.
You can use the following general syntax to validate input:
input_to_validate = input(message_on_input)
while not is_valid(input_to_validate):
print(message_on_error)
input_to_validate = input(message_on_input)
where is_valid() your method for determining if the input is valid or not
Before coding anything please do learn the language properly and do look up proper syntax conventions. The following code is much more readable and conforms more to coding standards:
import random
male = random.choice([True, False])
num = random.choice(["1", "0"])
name = "joe"
def is_valid(password):
return password in ["joey", "loki"]
user_input = input("insert Password here: ")
# this while-loop runs as long as the user_input is not valid
while not is_valid(user_input):
print("Wrong Password, please try again.")
user_input = input("insert Password here: ")
# this part of the code is reached only when user_input is valid
if male:
print("hello")
else:
print("welcome")
if name == "joe":
if num == "1":
print("hi world")
else:
print("nice")
if name + num == "joe0":
print("that's working")
Related
I'm trying to figure out how to clear an "if" condition and how to fix the result = print(x) part of my code. I'm trying to create a little search code based on the variable data, but I can't figure a few things out:
import time
def start():
data = ["Lucas_Miguel", "João_Batista", "Rafael_Gomes", "Bruna_Santos", "Lucas_Denilson"]
print("1" + " - Check Name")
print("2" + " - Register a New Name")
option = input("Choose an option: ")
if option == "1":
def other():
name = input("Type the first name: ")
for x in data:
if name in x:
result = print(x)
while True:
print("Yes " "or " "No")
confirm = input("Is the name you want in the options?: ")
if confirm == "Yes":
break
if confirm == "No":
print("Yes", " or", " No")
try_again = input("Do you want to write again?: ")
if try_again == "Yes":
return other()
other()
else:
print("Option not available")
time.sleep(1)
return start()
start()
The first problem is in the result = print(x) part. It works, but when the answer is more than one name, only the first one appear and I don't know how to fix it.
The second problem is in the "confirm = input" part. Basically, if the person answered with "No", when they go back, the answer will still be saved and the input will run twice, the first time with the saved answer and the second with the new answer. So I want to be able to clear that before the person answer it again.
I want to apologize already if the code is ugly or weird, but I started a few days ago, so I'm still learning the basics. Also thanks in advance for the help.
There is quite a bit here to unpack and like the comment on the question suggests you should aim to look at how to ask a more concise question.
I have some suggestions to improve your code:
Split the other into its own function
Try to use more accurate variable names
As much as you can - avoid having multiple for loops happening at the same time
Have a look at list comprehension it would help a lot in this case
Think about whether a variable really belongs in a function or not like data
What you're asking for is not immediately clear but this code should do what you want - and implements the improvements as suggested above
import time
data = ["Lucas_Miguel", "João_Batista", "Rafael_Gomes", "Bruna_Santos", "Lucas_Denilson"]
def other():
name_input = input("Type the first name: ")
matches = [name for name in data if name_input in name]
if len(matches) == 0:
print ("No matches")
for name in matches:
print(name)
while True:
print("Yes " "or " "No")
confirm = input("Is the name you want in the options?: ")
if confirm == "Yes":
break
if confirm == "No":
print("Yes", " or", " No")
try_again = input("Do you want to write again?: ")
if try_again == "Yes":
return other()
else:
return
def start():
print("1" + " - Check Name")
print("2" + " - Register a New Name")
option = input("Choose an option: ")
if option == "1":
other()
else:
print("Option not available")
time.sleep(1)
return start()
start()
The first problem will be solved when you remove 8 spaces before while True:.
The second problem will be solved when you add return (without arguments) one line below return other() at the indentation level of if try_again == "Yes":
Everybody can see that you are just learning Python. You don't have to apologize if you think, your code is "ugly or weird". We all started with such small exercises.
I am new to coding and I am learning python. I’m trying to write a simple program to test my skills, but I’m having some difficulties with it; I want to turn it into a function in order to make the program cleaner, but I get this error: http://prntscr.com/im5pt7
Here is what I want to put inside a function:
name = input(str("\nFull Name: "))
position = input(str("Position at the company: "))
print("\nConfirm Staff Data:\n")
name_confirm = "Name: %s"%(name)
position_confirm = "Position: %s"%(position)
print(name_confirm)
print(position_confirm)
confirmAns = input("Is the information right? (Y/N)")
if confirmAns == "y" or confirmAns == "Y":
message = "\nSearching for %s"%(name)
print(message)
hoursWorked = int(input("Insert hours worked: "))
if hoursWorked <= 0:
print("Please insert a valid number")
elif hoursWorked > 0:
print("\nCalculete Paycheck")
hourRate = int(input("Insert the rate of each hour worked: "))
bonus = input("If a bonus was given insert it here: ")
fine = input("If a fine was given insert it here: ")
print('\n')
payment = hoursWorked*hourRate-int(fine)+int(bonus)
paymentMsg = "Your Payment is: $%d"%(payment)
print(paymentMsg)
elif confirmAns == "n" or confirmAns == "N":
ctypes.windll.user32.MessageBoxW(0, "The software will close to avoid slowness.", "Warning", 1)
else:
print("Please answer with Y or N")
I've tried this but it did not work.
Here is all the code (working but with out the function so I need to copy and paste code): https://pastebin.com/PA9mxMkk
What is happening is that the function as other statements needs to hold it's code into a new indentation level
print('a')
def test(var):
print(var)
not this way
print('a')
def test(var):
print(var)
because this way it will give you the error that you are seeing.
All python code should be indented after the ':' character, in python the indentation should be 4 spaces, or people use the tab key, your code has an issue with indentation which I can't be bothered finding;
for example a 'class'
class this_is_a_class():
#indentation
#code goes here
pass
or a 'for loop' or 'while loop';
numbers = [0,1,2,3,4,5,6,7,8,9]
for number in numbers:
#indentation
print(number)
x = 0
while x < 10:
#indentation
x += 1
print('This is going to print 10 times')
or an 'if statement';
true_boolean = True
if true_boolean:
#indentation
print(True)
or a 'function';
def function():
#indentation
print('You have called a function')
What is actually happening, is python is reading through your code 'Token' by token and 'interpreting' what your code does. But considering you don't know what a function is; gloss over this paragraph.
Now for your question about how functions work. A function is used organize code. You can call a function multiple times, which makes your code more organized and easier to work with, this is why as your code got longer, you ran into this problem; Lets for example say i wanted to print 'hello world' 10 times.
I could write this code on 10 seperate lines;
print("hello world")
print("hello world")
#etc... More chance for error
or I could use a function and call it 10 times;
def say_hello_world():
#indentation
print("hello world")
for each_call in range(0,10):
say_hello_world() #This is the function call
You can also pass 'arguments into a function' for example;
def say_hello(person):
#indentation
print('hello', person)
say_hello('Alex')
Now any words that are in quotations in this answer can be google searched with the word 'python' and you can find out much more about how python works.
I hope this gets you started with python. Although all of these concepts can be used in other programming languages.
The first step which is often difficult in learning python in understanding indentation.
for example.
def hello_world(world):
print("hello ", world)
#your function code goes here.
#you need to indent back to be out of function block.
hello_world("there!")
out: hello there
so in your case it should be like this.
def AnsNo():
name = input(str("\nFull Name: "))
position = input(str("Position at the company: "))
print("\nConfirm Staff Data:\n")
name_confirm = "Name: %s"%(name)
position_confirm = "Position: %s"%(position)
print(name_confirm)
print(position_confirm)
confirmAns = input("Is the information right? (Y/N)")
if confirmAns == "y" or confirmAns == "Y":
message = "\nSearching for %s"%(name)
print(message)
hoursWorked = int(input("Insert hours worked: "))
if hoursWorked <= 0:
print("Please insert a valid number")
elif hoursWorked > 0:
print("\nCalculete Paycheck")
hourRate = int(input("Insert the rate of each hour worked: "))
bonus = input("If a bonus was given insert it here: ")
fine = input("If a fine was given insert it here: ")
print('\n')
payment = hoursWorked*hourRate-int(fine)+int(bonus)
paymentMsg = "Your Payment is: $%d"%(payment)
print(paymentMsg)
elif confirmAns == "n" or confirmAns == "N":
print("working")
else:
print("Please answer with Y or N")
return
I need help with a one or two things on my coursework. firstly when I input an incorrect username/password it will output more than one error message at a time. Secondly, when the men function ends the program will go back to the login loop( that allows the user to re-enter their username/password), I have fixed this by using the B variable but this feels extremely janky and I know there is a better way of doing it. Any Modifications or Tips are greatly appreciated (even if they're not related to my above problems). also the way i code is a bit weird so it's probs best you read from bottom to top :)
def quiz():
print ("quiz")# place holder for real code (let's me know the function has been run)
def report():
print ("report") # place holder for real code (let's me know the function has been run)
def a_menu():# Admin menu
print ("Welcome to the student quiz!")
print ("Please choose a option")
while True:
print ("1: Computer science")
print ("2: Histroy")
print ("3: Reports")
menu_in = int(input())
if menu_in == 1:
Comp_sci = True
quiz()
break
elif menu_in == 2:
Hist = True
quiz()
break
elif menu_in == 3:
report()
break
else:
print ("Invalid input! Please try again..")
def menu():
print ("Welcome to the student quiz!")
print ("Please choose a quiz")
while True:
print ("1: Computer science")
print ("2: Histroy")
menu_in = int(input())
if menu_in == 1:
Comp_sci = True
quiz()
break
elif menu_in == 2:
Hist = True
quiz()
break
def login():
b = False
print ("Please enter your username and password")
while True:
if b == True:
break
log_user = input("Username:")
log_pass = input ("Password:")
with open("D:\\Computer science\\Computing test\\logindata.txt","r") as log_read:
num_lines = sum(1 for line in open('D:\\Computer science\\Computing test\\logindata.txt'))
num_lines = num_lines-1
for line in log_read:
log_line = log_read.readline()
log_splt = log_line.split(",")
if log_user == log_splt[0] and log_pass == log_splt[2]:
if log_splt[5] == "a": # Admin will have to mannually change the admin status of other user's through .txt file.
b = True
a_menu()
else:
b = True
log_read.close()
menu()
break
else:
print ("Incorrect username or password, please try again!")
def register():
print ("enter your name, age, and year group and password")
while True:
reg_name = input("Name:")
reg_pass = input ("Password:")
reg_age = input ("age:")
reg_group = input ("Year Group:")
print ("Is this infomation correct?")
print ("Name:",reg_name)
print ("password:",reg_pass)
print ("Age:",reg_age)
print ("Year Group:", reg_group)
reg_correct = input ("[Y/N]").lower()
if reg_correct == "y":
reg_user = reg_name[0:3]+reg_age
reg_write = open("D:\\Computer science\\Computing test\\logindata.txt","a")
reg_write.write (reg_user+","+reg_name+","+reg_pass+","+reg_age+","+reg_group+","+"u"+"\n")
print ("Your username is:",reg_user)
reg_write.close()
login()
break
elif reg_correct == "n":
print ("Please Re-enter your infomation")
else:
Print ("Invalid input! Please try again...!")
def login_ask():
print ("Do you already have an account?")
while True:
ask_in = input("[Y/N]").lower()
if ask_in == "y":
login()
break
elif ask_in == "n":
register()
break
login_ask()
I'm writing a very basic payment system on Python, but every time I enter the main() function and input a value it just quits the program. Any ideas?
import sys
def info():
usernames = ["15mermelstein_p","gril","jeff","example"]
passwords = ["test","pass","what","why"]
balances = [22.91,45.76,21.62,7.54]
a = input("What is your username? ").lower()
b = input("What is your password? ")
print(a)
print(b)
success = 0
for index in range(len(usernames)):
if usernames[index] == a and passwords[index] == b:
user_num = index
success = 1
break
if success == 1:
print("Login successful\n")
main()
return (a,b,user_num)
else:
print("Login unsuccessful")
info()
def main():
choice = input("-----Welcome to the new School Payment System-----\n1. Add money to account\n2. Change password\n3. Change your username\n4. Quit the program\n--------------------------------------------------\n")
if choice == "1":
credit = input("How much money would you like to deposit into your account? ")
temp_b = input("Please type in your password once againto confirm thios transaction: ")
if temp_b == info[2]:
balances[num(info[3])] += float(credit)
else:
print("The password you entered was incorrect, please return and try again. ")
elif choice == "2":
check_pass = input("Please input your current password first")
elif choice == "3":
sys.exit(0)
elif choice == "4":
sys.exit(0)
else:
sys.exit(0)
info()
Since you have provided no other information and the code runs fine on my machine, I am going to assume that your error is that you are running the wrong version of python. The code compiles with python-2.x but when you get to any of the inputs, it will not work as desired:
AJs-MacBook-Pro:~ $ python2 k.py
What is your username? "gril"
What is your password? "pass"
gril
pass
Login successful
-----Welcome to the new School Payment System-----
1. Add money to account
2. Change password
3. Change your username
4. Quit the program
--------------------------------------------------
1
AJs-MacBook-Pro:~ $
I ran your code with python3.x and it has some errors.
if temp_b == info[2]:
balances[num(info[3])] += float(credit)
You can subscript a function object. What you really need to do is either pass the correct username password to the main function so that it can be accessed in the main function to add balance and other things and validate the password again.
First of all, your program works fine apart from the below code,
if temp_b == info[2]:
balances[num(info[3])] += float(credit)
You are trying to access info as an array, but info is a function. (may be you have missed to define array for your payment system menu options).
I have a problem with a condition in my python code.
It's a mathematics application, and here's the part of the code that is not working well:
def askNumber():
"""Asks the number to test"""
a=raw_input("Select the number to test (type 'exit' for leaving):")
if len(a)!=0 and a.lower!="exit":
try:
b= int(a)
processing(b)
except ValueError:
print "Your input is not valid. Please enter a 'number'!"
time.sleep(1)
askNumber()
elif len(a)!=0 and a.lower=="exit":
answer()
else:
print "Your input can't be 'empty'"
time.sleep(1)
askNumber()
So, when in the raw_input for "a" I type "exit", the supposed condition to apply is the elif one but ends up applying the if one, ending up printing "Your input is not valid. Please enter a 'number'!" Sorry, if it's something obvious, I'm a begginer, although I tried to find the mistake several times.
You need to call the .lower() function.
if len(a) != 0 and a.lower() != "exit":
# ...
elif len(a) != 0 and a.lower() == "exit":
There is no real need to test for len(a)!=0, simply test for a itself:
if a and a.lower() != "exit":
# ...
elif a and a.lower() == "exit":
Empty strings evaluate to False in a boolean context.
Your program flow is a bit inside out, may I suggest some improvements?
def askNumber():
"""Asks the number to test"""
while True:
a = raw_input("Select the number to test (type 'exit' for leaving):")
if not a:
print "Your input can't be 'empty'"
continue
if a.lower() == "exit":
answer()
break
try:
b = int(a)
except ValueError:
print "Your input is not valid. Please enter a 'number'!"
continue
processing(b)
Actually, the not a branch can be eliminated as well (empty inputs will be handled in except).
You could change the condition for the following one:
if a and a.lower() !="exit":
# .....
elif a and a.lower() == "exit":
answer()
elif a and not a.isdigit(): print "invalid input"
else:
#.............
Please note that yo don't need len(a) != 0 , just by using a will evaluate if it's empty or not.