Python: Save a hex string to a jpg image file

Question

I have multiple hex strings like this one (big string so I truncated the middle):

0xFFD8FFEE000E41646F626500640000000002FFE11E2445786966000049492A006A0500002C010000010000002C0100000100000041646F62652050686F746F73686F702043533620284D6163................................................................................................................................................................7D8D9DAE1E2E3E4E5E6E7E8E9EAF1F2F3FF7F8F9FAFFC4001F0100030101010101010101010000000000000102030405060708090A0BFFC400B511000201020404030407050404000102770

I want to save each string in a file. But my code is not working well:

import binascii

data = binascii.a2b_hex(my_hex_string)
with open('/path/image.jpg', 'wb') as image_file:
    image_file.write(data)

I receive this error:

TypeError: Odd-length string

When I remove the first 0 I get this:

TypeError: Non-hexadecimal digit found

When I remove the two first chars (since JPG is supposed to start with FF D8), I get this again:

TypeError: Odd-length string

Any ideas please?

Answer 1

Are you sure the string doesn't have something extra in it? Whitespace, newlines, etc?

Try my_hex_string.strip()

Also, there is possibility, you can have spaces inside the string, so you can do something like that to clean them out:

binascii.a2b_hex(toSend.replace(' ', ''))

Answer 2

3.5 yo thread. But is getting many views, so I will add my 2c. OP's code works just fine for me. My guess is that the hex string is corrupted. Most prominantly, not seeing an FFD9 for "end of image."

Answer 3

I had the same problem and I tested your code. I think there is nothing wrong with your code. Just try to remove the 0x from the beginning of your Hex Codes. I think that would work for you.