Break ecb

ecb.py

import requests
import math

flag = []
bytestocrunch = 32 # Get it with a block_size long pattern. bytes to crunch will be the (responce - (block_size*2))/2. *2 because we get a hexstring
block_size = 16 # Get it by appending n bytes until the output increases. Ater it increased count the bytes until it increases again
apattern = [0x61] * math.ceil(bytestocrunch/block_size) * block_size # Any byte will be sufficient
for y in range(1, bytestocrunch+1):
    pattern = ''.join([hex(b)[2:].zfill(2) for b in apattern[:-y]]) # Convert 0:-y to a hex string
    
    flag_enc = requests.get('http://aes.cryptohack.org/ecb_oracle/encrypt/{}/'.format(pattern)).json()['ciphertext']
    reference = flag_enc[0:len(apattern*2)]
        
    # Test every possible byte (2^8 bits)
    for x in range(0, 256):
        testbyte = hex(x)[2:].zfill(2) # Convert byte to hexstring
        testpattern = pattern + ''.join([hex(b)[2:].zfill(2) for b in flag]) + testbyte
        testenc = requests.get('http://aes.cryptohack.org/ecb_oracle/encrypt/{}/'.format(testpattern)).json()['ciphertext']
        if(testenc[0:len(apattern*2)] == reference):
            flag.append(int(testbyte, 16))
            break
print(''.join([chr(n) for n in flag]))