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python program, want to print 5 attempts done, if used

I want to print 5 attempts completed if i use all the attempts incorrectly
def user_name_repeat():
for repeat_user_name in range(4):
re_enter_user_name = input("Incorrect username, try again: ")
if re_enter_user_name == "vishesh":
print("Correct user name ur logged in")
break
input_ = input("enter user name: ")
if input_ == "vishesh":
print("Correct user id, you are logged in: ")
elif input_ != "vishesh":
print(user_name_repeat())

Why not wrap the whole thing inside a loop?
for _ in range(5):
input_ = input("enter user name: ")
if input_ == "vishesh":
print("Correct user id, you are logged in")
break
else:
print('Incorrect username, try again')
else:
print('Attempts exhausted')
More info on for/else
Edit: You could also fix up the logic in your own code with something like-
def user_name_repeat():
for repeat_user_name in range(4):
re_enter_user_name = input("Incorrect username, try again: ")
if re_enter_user_name == "vishesh":
print("Correct user name ur logged in")
return True
return False
input_ = input("enter user name: ")
if input_ == "vishesh":
print("Correct user id, you are logged in: ")
elif input_ != "vishesh":
if not user_name_repeat():
print('Attempts exhausted')
Really, all you need is to return an indicator (a boolean value in this case) from your function to let the caller know, whether the attempts succeeded.

Just use the else in your for loop, else will run after the loop is completed without encountering a break statement. Also, use return in your user_name_repeat() function, since you are calling it in print(user_name_repeat()) So it should return somethinig to the print() to print for:
def user_name_repeat():
for repeat_user_name in range(4):
re_enter_user_name = input("Incorrect username, try again: ")
if re_enter_user_name == "vishesh":
return("Correct user name ur logged in")
break
else:
return "5 attempts completed"
input_ = input("enter user name: ")
if input_ == "vishesh":
print("Correct user id, you are logged in: ")
elif input_ != "vishesh":
print(user_name_repeat())

Since you're working with a function, you should simply return the string instead of printing it inside of the function. That way, you can just return the "incorrect" string after the for loop.
def user_name_repeat():
for repeat_user_name in range(4):
re_enter_user_name = input("Incorrect username, try again: ")
if re_enter_user_name == "vishesh":
return "Correct user name ur logged in"
return "Number of incorrect tries exceeded limit"
Another approach would be using the else clause of the for loop. In loops, the code inside the else clause will only be executed if (and when) the loop ends "naturally"; that is - when the condition running the loop stops being valid (when the for reaches the end of the range(4), for example).
In your code, the loop only ends if the limit of incorrect attempts is reached, since you break the loop if the user inputs the right password. So the following could be done:
def user_name_repeat():
for repeat_user_name in range(4):
re_enter_user_name = input("Incorrect username, try again: ")
if re_enter_user_name == "vishesh":
print("Correct user name ur logged in")
break
else:
print("Number of incorrect tries exceeded limit")
However, I should point out that the first option is way better, specially considering you're printing the return of the function call later (and if the function has no explicit return, you're actually printing None).
You can even take the code out of the function, which should make things easier for you, as Chase's answer points out ;).

How to make my code break out of loop and actually detect a certain input

I am doing a basic Python assessment on a password vault but am getting errors that I can't seem to solve myself. and btw guys I didn't come here for a grammar and punctuation lesson, so if you are just here to edit my question and not offer any assistance please dont bother.
For example, in this part of the code I want the user to input 1 or 2, and if he selects 1 it asks him to log in, while if he selects 2 it asks him to register. But at the moment it is completely ignoring the parameters and accepting anything.
Another problem is that when the user enters the valid password, instead of just stopping at password correct, it for some reason re-asks "what is your username."
while True:
login_orsignup1 = input('''Press
1) to Log in
2) to register a new account
''')
if login_orsignup1 != 1:
while True:
username = input('''What is your,
Username: ''')
if input_username == username:
l_p = input('''What is your password ''')
while True:
if l_p == input_lockerpassword:
print("Password Correct")
break
login_originup1()
----------------------------------------------------------#Full code begins now
l_p = ""
print("------------------------------------------------------------------------")
print('''Welcome to password Locker, a place where you can
store all your passwords to easily enter your precious accounts without
hassle.''')
print("------------------------------------------------------------------------")
print('''First lets make an account,''')
while True:
first_name = input('''What is your first name?
''')
if first_name.isdigit(): #isdigit, detects if there
print("Please enter a valid answer, No nubers shoud be present")
elif first_name == "":
print("Please enter an answer")
#the continue code skips the boundries within the loop and carries on with the connected program until it is succesfully met
else:
break #the break loop exits the current loop and continues with the next programes following it
while True:
sur_name = input('''What is your surname?
''')
if sur_name.isdigit(): #isdigit detects if the
print("No numbers")
elif sur_name == "":
print("Please enter an answer")
#the continue code skips the boundries within the loop and carries on with the connected program until it is succesfully met
else:
break
print('''------------------------------------------------------------------------''')
print('''Welcome, {} {}
what would you like your username to be, it should be something
memorable and no longer than fifteen characters long, '''.format(first_name, sur_name))
while True:
input_username = input("")
if 0 < len(input_username) < 16:
print('''Nice, username''')
break
elif input_username == "":
print("Please enter an answer")
else:
print('''Your username should be a maximum of 15 charecters, ''')
print('''-------------------------------------------------------------------------''')
while True:
input_lockerpassword = input('''Now it's time to setup a password for your locker, It should be between 4
and 10 charecters long,
''')
if len(input_lockerpassword) > 4 and len(input_lockerpassword) < 11:
print('''{}, is locked in thanks for joining Password Locker'''.format(input_lockerpassword))
break
else:
print("It should be between 4 and 10 charecters long!")
print('''
-------------------------------------------------------------------------------------------''')
def login_originup1():
print(''' Welcome to password vault, You can either login or create a New account''')
while True:
login_orsignup1 = input('''Press
1) to Log in
2) to register a new account
''')
if login_orsignup1 != 1:
while True:
username = input('''What is your,
Username: ''')
if input_username == username:
l_p = input('''What is your password ''')
while True:
if l_p == input_lockerpassword:
print("Password Correct")
break
login_originup1()```

Ok, first of all, you should know that the input() function returns a string and, as such, your first condition : if login_orsignup1 != 1 will always be true, because the string object '1' isn't equal to the int object 1. As for why you get asked again for the user after having a good password, that is because the break statement only breaks from the current loop. So you only break of this loop to get back at the start of your username verification loop. I would suggest a cleaner implementation like so :
# login or sign-up loop
while True:
login_orsignup1 = input(" Press \n1) to Log in \n2) to register a new account")
# you can have the input result in a variable like so, if you want to use it later on
if login_orsignup1 == "1": # you said '1' was for login, right?
# or you can test this directly in your if statement
# if it is only needed for this condition
while input("What is your username: ") != username:
print("Incorrect username")
# same thing for password, but could be implemented another way if you
# don't want to loop until the user gets the correct password
while input("What is your password: ") != input_lockerpassword:
print("Incorrect password for this username")
# your code gets here only if the user entered the right credentials
# so you can now break of the login or sign-up loop
break
elif login_orsignup1 == "2":
# code for registration here
This could be good enough for a simple thing. I would recommend designing this console program by following concepts of a state-machine and adding more code at each step to handle cases like going back one step or back at the start.
Hope this helps

the problem is in your login_originup1 function you are making three While loops that the program can't escape from in your function you are asking if login_orsignup1 != 1
without an else statement so if the user wanted to login he would press input in "1" then the program will say that
"1" =! 1 is false
it will look for an else statement But not find one so it will go back to the start of the loop and ask the user to input again. this is it for the First Loop.
Now if the user Inputs in "2" (which means that the user wants to register) it will make him log-in because:
"2" =! 1is true
and will continue to the next while loop in here you will be asking for the username and the user will give the username. Now this is it for the Second Loop
we now go to the last loop where you ask for the Password and the User Will give the Password. The program Will either 1. say that it was false and ask for the password again or 2. it will accept the password and Break the While loop. Now this is it for the Third Loop
so why is it asking me for the Username Because the break statement breaks only the while loop it is in so that break statement broke only the third while loop and was back to the Second Loop which the Second Loop will bring us back into the Third Loop again
so how to fix this?
simple like this:
def login_originup1():
print('Welcome to password vault, You can either login or create a New account')
while True:
login_orsignu = input('Press\n1) to Log in\n2) to register a new account\n')
loopBreaker = 0
if loopBreaker:
break
if int(login_orsignu) != 1:
while True:
if loopBreaker:
break
username = input('What is your,\nUsername:')
if input_username == username:
l_p = input('What is your password ')
while True:
if loopBreaker:
break
if l_p == input_lockerpassword:
print("Password Correct")
loopBreaker = 1
break

How to prevent user from inputting spaces/nothing in Python?

I have a problem in which users can input spaces or nothing and still pass through the program, how do I go about preventing this? I am still a beginner at python.
def orderFunction(): # The function which allows the customer to choose delivery or pickup
global deliveryPickup
deliveryPickup = input("Please input delivery or pickup: d for delivery p for pickup")
if deliveryPickup == "d":
global customerName
while True:
try:
customerName = (input("Please input your name"))
if customerName == (""):
print("Please input a valid name")
else:
break
global customerAddress
while True:
try:
customerAddress = (input("Please input your name"))
if customerAddress == (""):
print("Please input a valid Address")
else:
break
global customerPhnum
while True:
try:
customerPhnum = int(input("Please input your phone number"))
except ValueError:
print("Please input a valid phone number")
else:
break
print("There will also be a $3 delivery surcharge")
elif deliveryPickup == "p":
customerName = (input("Please input your name"))
if customerName == (""):
print("Please input a valid name")
orderFunction()
else:
print("Please ensure that you have chosen d for Delivery or p for Pickup")
orderFunction()
orderFunction()
Here is my attempt at doing this but I get all kinds of unindent and indent errors at the moment and I think my while loops are probably wrong.
Essentially if I input a space or hit enter into one of the customer inputs (customerName for instance) it gets stored. This needs to prevented and I have tried to fix it by using while loops which obviously haven't worked.
Hopefully someone has a solution to this problem
Many Thanks.

.strip() removes all tabs or spaces before and after a string.
Meaning all spaces == empty string. All tabs == empty string. So all you have to check if the length of that string != 0 or the string is not empty. Just use an infinite loop to keep on forcing the right input.
Also as a tip, you don't have to limit yourself into one function.
Here's a working code below.
def getNonBlankInput(message, error_message):
x = input(message)
while len(x.strip()) == 0:
x = input(error_message)
return x
def getValidIntegerInput(message, error_message):
msg = message
while(True):
try:
x = int(input(msg))
break
except ValueError:
msg = error_message
return x
def orderFunction(): # The function which allows the customer to choose delivery or pickup
global deliveryPickup
global customerName
global customerAddress
global customerPhnum
deliveryPickup = input("Please input delivery or pickup: d for delivery p for pickup")
if deliveryPickup == "d":
customerName = getNonBlankInput("Please input your name: ", "Please input a valid name: ")
customerAddress = getNonBlankInput("Please input your address: ", "Please input a valid address: ")
customerPhnum = getValidIntegerInput("Please input your phone number: ", "Please input a valid phone number: ")
print("There will also be a $3 delivery surcharge")
elif deliveryPickup == "p":
customerName = getNonBlankInput("Please input your name: ", "Please input a valid name: ")
else:
print("Please ensure that you have chosen d for Delivery or p for Pickup")
orderFunction()
orderFunction()

Try using a regular expression that checks if any character between "A-Z" has been inserted, if not, give an error

The while loops are a decent solution, you just need to add more checks to your if statements.
First, you don't need a try statement on the top two loops. Don't use a try statement unless you're expecting an error, which you need to handle with an except statement, like you do in the bottom while loop.
Then you just need to add more conditions to your top two loops, I don't know exactly what you want to prevent, but you could try checking the length of the input, also see this answer for an interesting method:
https://stackoverflow.com/a/2405300/8201979

Instead of using input right away you can make a function similar to this one that will only allow valid inputs.
You can use this valid_input function instead of input.
def valid_input(text):
not_valid = True
res = ''
while not_valid:
res = input(text)
if res.split(): # if text is empty or only spaces, this creates an empty list evaluated at False
not_valid = False
return res
here the check is pretty simple: every text made out of nothing or spaces won't be allowed and we will keep asking for the same input until a valid information is given.
I made this code simple just so you get a general idea. But you can change the validation test to your liking and maybe also output a warning saying why the input wasn't allowed so the person knows what to do. You can do more advanced validation with regex, and maybe you need a minimum text length etc...

You have indent error because you have a try statement without the corresponding except.
You need both to make it work (as you did in the Phone number section).
Here is a link to the try/except: docs
Also, you can check if a string is empty as detailed in this answer.
So for example you want to write:
try:
customerName = input("Please input your name")
if not customerName:
print("Please input a valid name")
else:
break
except ValueError:
print("Please input a valid name")
Although the above seems a bit redundant, so you might want to raise an exception if the customer name is empty, catch the exception in the except block, print the warning and return error (or something else).
try:
customerName = input("Please input your name")
if not customerName:
raise ValueError
except ValueError:
print("Please input a valid name")
else:
break

Try adding another while true for pick and delivery option so that it can prevent taking other inputs

you don't need any of those try/excepts (which are broken anyway).
Its difficult to figure out what you're trying to do, are you trying to raise an exception if an empty string is passed, or request another input from the user? You seem to be half implementing both at the moment.
If its the latter, something like this would work.
def func(fieldname):
while True:
val = input("Please input your {}".format(fieldname))
if val.strip() != "":
break
else:
print("Please input a valid {}".format(fieldname))
return val
delivery_pickup = input("Please input delivery or pickup: d for delivery p for pickup")
if delivery_pickup == "d":
customer_name = func("name")
address = func("address")
phone_number = func("phone number")

What you are looking for is the str.strip method that remove trailing whitespace in strings.
Also I think try is not particularly suited for your needs here.
customerName = input("Please input your name")
while not customerName.strip():
customerName = input("Please input a valid name")
for the phone number I would not convert to integer because if the phone number starts with zeros, they will not be stored.

Use of continue in python - how to break out of an individual loop?

I have written some code that contains several while loops.
answer = "yes"
while answer == "yes":
name = input("what is your name?")
while len(name) > 0 and name.isalpha():
print("okay")
break
else:
print("error")
continue
job = input("what is your job?")
while job.isalpha():
print("great")
else:
print("not so great")
continue
age = input("how old are you?")
while age.isnumeric():
print('nice')
else:
print("not so nice")
continue
What I want the code to do is check for a valid name entry, and if it is invalid re-ask the user for their name. I want it to do the same thing with their job. Unfortunately, when I use continue after the job statement, instead of just re-asking for their job, it re-asks for their name as well. Does anyone know how to make it only re-ask for the job, and not re-do the whole program? Does anyone know how I could replicate this for multiple while loops I.e. Asking for job, name, age, star sign etc.?
Thank you.

You can use while True because answer never change.
It's useless while for 1 expression and 1 break.
You should put everything on a while True and check:
while True:
name = input("what is your name?")
job = input("what is your job?")
if len(name) > 0 and name.isalpha() and job.isalpha():
print("great")
break
print("error")

I moved the question into a seperate fuction to de-duplicate the code. It will ask again and again as long as the answer is not satisfactory.
def ask(s):
r = ""
while not r or not r.isalpha():
r = input(s)
return r
name = ask("What is your name?")
print("Okay. Your name is {}.".format(name))
job = ask("What is your job?")
print("Great. Your job is {}.".format(job))

My Python function 'loops' for no reason

I'm writing a Python function to collect user input and then return it as a dict.
The input should meet certain criteria, so I check for that and I ask the user to supply his input again if it doesn't check out.
This is my code:
def getIDs(stageid=0):
if stageid == 0:
stageid = input("Please supply stageID: ")
if (len(stageid) < 4 or len(stageid) > 5) or not stageid.isdigit():
print("Wrong stageID, try again.")
getIDs()
raceid = input("Please supply the raceID: ")
if (len(raceid) < 1 or len(raceid) > 3) or not raceid.isdigit():
print("Wrong raceID, try again.")
getIDs(stageid)
else:
# print(stageid)
return {"stageid": stageid, "raceid": raceid}
What I do is have the function just call itself if the stageid isn't correct, but I don't want the user to have to re-enter the stageid again if the raceid is incorrect, so if the raceid is incorrect, the I call the function with the stageid as a parameter.
My problem is, however, that if I get the stageid wrong several times, then get it right and continue to the raceid, it will ask for the raceid for as many times as I had entered the stageid. I can't figure out why. The commented print function prints all supplied stageid's in reverse order and in the end, the return value contains the initially supplied (wrong) stageid and the raceid.
Does anybody know why Python somehow loops through all values?
Vasilis

Just add a return statement when calling the recursive function so the termination will exit from that instance.
def getIDs(stageid=0):
if stageid == 0:
stageid = input("Please supply stageID: ")
if (len(stageid) < 4 or len(stageid) > 5) or not stageid.isdigit():
print("Wrong stageID, try again.")
return getIDs()
raceid = input("Please supply the raceID: ")
if (len(raceid) < 1 or len(raceid) > 3) or not raceid.isdigit():
print("Wrong raceID, try again.")
return getIDs(stageid)
else:
# print(stageid)
return {"stageid": stageid, "raceid": raceid}