固定斐波那契数从 1 到 50 Python

Question

I have a set of numbers from 1 to 50 and from these numbers I have to find which are Fibonacci number, after that I have to find which of these numbers contain the number 1 and 2.

For example in my code my Fibonacci numbers are 0, 1, 2, 3, 5, 8, 13, 21, 34 I have written a code, but I can't add the last part of the question, so it finds out which of these numbers contain a number 1 or 2 in them. here is my code

def isPerfectSquare(x):
    s = int(math.sqrt(x))
    return s * s == x


def isFibonacci(n):
    return isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4)


for i in range(0, 51):
    if isFibonacci(i) == True:
        print(i, "is a Fibonacci Number")
    else:
        print(i, "is a not Fibonacci Number ")


for q in range(1, 51):
    if isFibonacci(i) == True and '1' in i and '2' in i:
        print(q)
    else:
        print('Error') 

the result is something like this is gives me all of these numbers 0, 1, 2, 3, 5, 8, 13, 21, 34 as Fibonacci and the rest are not which is perfect but then it keeps giving me error, what i wanted to do is it to write a loop that contains all Fibonacci numbers like these 0, 1, 2, 3, 5, 8, 13, 21, 34 and that contain a 1 and 2 in the number to be printed out but it just prints everything as Error.

Answer 1

for q in range(1, 51):
    if isFibonacci(q) == True and '1' in str(q) and '2' in str(q):
        print(q)
    else:
        print('Error') 

1. This loop's variable is q , not i .
2. You have to convert the number to string using str(...) .