CODEWITHSUNDEEP
arrayspython-3.xregex
magicsword
magicsword

Reputation: 1289

how to get the next 2 bytes from a bytearray after a match

I have a bytearray:

s = b'\x01\x80\x00\x04_\xa9\xa20\x01\x19\x00'

I want to search and find:

\xa9\xa20

Then once found I want to get the next 2 bytes after that. In this case:

\x01\x19

I've tried:

m = re.search(b'[(?:\xa9\xa20)]{2}',s,re.DOTALL).group(1)

But get an Index error:

m = re.search(b'[(?:\xa9\xa20)]{2}',s,re.DOTALL).group(1)
IndexError: no such group

Upvotes: 1

Views: 78

Answers (2)

Alireza
Alireza

Reputation: 2123

Try this pattern:

\\xa9\\xa20(\\xa?\d+\\xa?\d+)

See Demo in Regex101

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 627077

You have a corrupt regex pattern here since you put a (?:\xa9\xa20) non-capturing group into a character class (repeated twice). However, just removing the square brackets won't help.

You can fix the code using

import re
s = b'\x01\x80\x00\x04_\xa9\xa20\x01\x19\x00'
m = re.search(b'\xa9\xa20(.{2})', s, re.DOTALL)
if m:
    print(m.group(1)) # => b'\x01\x19'

See the Python demo

That is, rather than unsuccessfully trying to match two consecutive occurrences of \xa9\xa20 bytes, match a \xa9\xa20 byte sequence once and then match and capture any two bytes after them with (.{2}) capturing group.

Upvotes: 1

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