Simple passcode checker

Question

OK, this is racking my brain. Ive stripped this code back down to the bare bones for this question. How do i get the code to only print "Valid Code" only once, meaning after it has checked all numbers? Example, if i type 12345, i dont want it to say "Valid Code" 5 times, only once. I did put a break in the "Invalid" part, because as soon as it sees 67890, there is no need to continue checking the rest of the numbers. Keep in mind, i fully understand why i get this output with the current "print" statements i have show. I have tried to put it at the end, but if i did an "Invalid" entry, it would still print "valid" after the code ran. This is nothing real, and only for my own learning. Thanks

https://code.sololearn.com/cpo3349NtS4w/?ref=app

Answer 1

You actually don't need to change your code much. What you desire can simply be obtained using a boolean.

https://code.sololearn.com/cGjTpwzLP71Q/?ref=app

Answer 2

def numbercheck():
code = input("Enter a number- ")
if all(n in '12345' for n in code):
print('Valid Code')
elif any(n in '67890' for n in code):
print('Invalid Code')

numbercheck()

Note that this won't output anything if you enter something like "fd f hjjh" and the elif statement is technically not necessary if the input is guaranteed to be a number

Answer 3

NUKE Yes, it is incorrect. The output should only have ONE statement - either "Valid code" or "Invalid code" - based on the condition.

Answer 4

NUKE It doesn't. It sees the first character ('4'), prints "Valid Code" and leaves the loop without checking any of the other characters

Answer 5

https://code.sololearn.com/c6jprV5PJ9et/?ref=app

Answer 6

NUKE , that would not work. If the first number is good, it breaks out and finishes. If you input 456, it would still show good even though a 6 is present.

Answer 7

Just don't use loop => try it with if-elif construction and string comparison. Please look at the code.
https://code.sololearn.com/cjfdbB80i6vK/?ref=app. Hope it helps you 😉

Answer 8

Ok Kainatic [#MakeYourMark]

Answer 9

It's tough

Answer 10

def numbercheck():
code = input("Enter a number- ")
for x in (code):
if x in ("12345"):
print("Valid Code")
break
#just place break here and u good bro!
elif x in ("67890"):
print ("Invalid Code")

numbercheck()

Answer 11

Anna I am getting
Valid code
Valid code
Invalid code for 456 input is it incorrect?

Answer 12

On typing 456 it does show invalid code
I think its okok Beau

Answer 13

you forgot to return to 0