Byte At A time ECB Attack This is my first post on stack overflow/stack exchange so please excuse me if I asked the question in a bad way or the format is bad.
I am currently working through the cryptopals challenges and I got stuck on this one: https://cryptopals.com/sets/2/challenges/12 where you basically need to figure out a secret by extracting one byte a time exploiting AES-ECB (You are given a function which encrypts usertext + secret and encrypts it with AES-ECB). I understand why this attack works but what I don't understand is how we can extract more than one block size worth of bytes of the secret. Because let's say for block size = 8. We do the attack and discover the first eight bytes of secret, how can we get all the other bytes of secret as whatever amount of bytes we will enter now only parts of the first block size of secret will get discovered.
You have and oracle call of
$$O(\texttt{your-string}) = (\operatorname{AES-128-ECB}(\texttt{your-string} || \texttt{unknown-string})$$
Since you know that the AES has 16-byte block cipher and it uses CBC, request;
$$C = O(\texttt{AAAAAAAAAAAAAAAx})$$ from oracle, where $\texttt{x}$ is the first byte of the random key.
Now ask to oracle  these encryptions $$C = O(\texttt{AAAAAAAAAAAAAAA[a..zA..Z]})$$ and see which character is equal to $\texttt{x}$, sat $1$
Now once we determine the $\texttt{x}$, we can move to the next one. Request from Oracle;
$$C = O(\texttt{AAAAAAAAAAAAAAxy})$$
Now ask to oracle these encryptions $$C = O(\texttt{AAAAAAAAAAAAAAx[a..zA..Z]})$$ and see which character is equal to the second byte $\texttt{y}$ of the string, say $2$
With this method, we can get 16 bytes of the secret. What about if there are more. Request
$$C = O(\texttt{AAAAAAAAAAAAAAA1})$$
then request this will provide the next block's encryption as the input
$$\texttt{234567890abcdef}$$, where you know $\texttt{234567890abcde}$ but not $\texttt{f}$, but you got the encryption of this block. Now ask the oracle
$$C = O(\texttt{234567890abcde[a..zA..Z]})$$
to determine the 17th byte $\texttt{f}$, the rest is similar.