Displaying the print where there is no error

Question

if password:
    if blank > 0:
        errorcount += 1
    if letter > 0:
       errorcount += 1
    if upper < 1:
        errorcount += 1
    if lower < 1:
        errorcount += 1
    if numbdig < 2:
        errorcount += 1
    if passwords < minimum:
        errorcount += 1
    print(password, 'is not valid', errorcount, 'errors!')
else:
    print(password, 'is valid!')

When there is 0 error the else should print but instead the not valid prints instead.

Answer 1

The output is correct because the case of no error means outer if block i.e. if password will be true, so your else will not be executed.

Inside if password, there is a print statement which is getting printed.

What you can do is, put a check on your errorcount variable & print accordingly:

if errorcount >= 1:
    print(password, 'is not valid', errorcount, 'errors!')
else:
    print(password, 'is valid!')