count = 5
while True:
UserName = input('Username Arabic Man pls:' )
PassWord = input('Dont be a nonce:')
Preset_Username = [{'42'},{'123'},{'preset'}]
Preset_Password = [{'hi'},{'yes'},{'maybe'}]
if UserName == Preset_Username and PassWord ==Preset_Password:
print('Welcome my friend, you are here to do what')
break
elif UserName != Preset_Username and PassWord != Preset_Password:
count -= 1
print('False input')
if count == 0:
break
Welcome to the forum. What sort of help are you looking for?
I’ve edited your post for readability. When you enter a code block into a forum post, please precede it with a separate line of three backticks and follow it with a separate line of three backticks to make it easier to read.
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- You don’t really want to store passwords in plain text. Check links like these for more info.
That said, if you just want a way of checking if input X and input Y match stored values for X and Y, there would be several ways. The way you’re approaching it, the dictionaries in your lists are just keys and not values and I don’t think that will work.
One, you could use two lists:
Preset_Username = [user1,user2,user3]
Preset_Password = [password1,password2,password3]
Then check to see if UserName = Preset_Username[i] and same Password (with same i value).
You could fix your syntax in the dictionaries in the lists, adding a key to each entry:
Preset_Username =[{"key":"value"},{"key":"value"}]
Then basically the same thing.
Or, combine them into a single list of dicts:
Preset_Info = [{"UserName1":"Password1"},{"UserName2":"Password2"}]
- But of course however you approach it you’ll need to loop through the list/keys. Your code now would only work if the user entered the value of the whole list as opposed to an entry:
Username =[{'42'},{'123'},{'preset'}]
Is the only way to make UserName == Preset_Username
Same issue with the password.
So, look into using for loops / if statements or other functions to loop through a list (and/or dicts within it) to check for a match. Try using just one input to just one list first, then adapt it for two lists / dicts / list of dicts.
