Python regex search for hexadecimal bytes

Python regex search for hexadecimal bytes

By : user106780
Date : November 17 2020, 11:58 AM
hop of those help? I'm not sure how to search the entire file and return all the matches. I'd like to return the offset in either decimal or hex
code :
import re

f = open('data.txt', 'wb')

f = open('data.txt', 'rb')
data = f.read()

pattern = "\xEB\xEC"
regex = re.compile(pattern)

for match_obj in regex.finditer(data):
    offset = match_obj.start()
    print "decimal: {}".format(offset)
    print "hex(): " + hex(offset)
    print 'formatted hex: {:02X} \n'.format(offset)

decimal: 2
hex(): 0x2
formatted hex: 02 

decimal: 6
hex(): 0x6
formatted hex: 06 

decimal: 10
hex(): 0xa
formatted hex: 0A 

decimal: 14
hex(): 0xe
formatted hex: 0E 

decimal: 18
hex(): 0x12
formatted hex: 12 

decimal: 22
hex(): 0x16
formatted hex: 16 

decimal: 26
hex(): 0x1a
formatted hex: 1A 

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Convert from string containing hexadecimal characters to bytes in python 3

Convert from string containing hexadecimal characters to bytes in python 3

By : Vineet
Date : March 29 2020, 07:55 AM
like below fixes the issue If all your codepoints are within the range U+0000 to U+00FF, you can encode to Latin-1:
code :
>>> '\xe8\x00\x00\x00\x00\x60\xfc\xe8\x89\x00\x00\x00\x60\x89'.encode('latin1')
Match hexadecimal notated bytes regex

Match hexadecimal notated bytes regex

By : user3065155
Date : March 29 2020, 07:55 AM
With these it helps Sorry to say that, but haxe EReg class misses some methods for frequent usecases. However, it's still achievable. Try this:
code :
class Test {
  static function main() {
    var sbytes: String = "cafebabe";

    var hexbyteRe = ~/[0-9a-f]{2}/;

    var pos = 0;
    while(hexbyteRe.matchSub(sbytes, pos)){
      var mp = hexbyteRe.matchedPos();
      pos = mp.pos + mp.len;
Python: split bytes with a hexadecimal delimiter

Python: split bytes with a hexadecimal delimiter

By : Gh0stExp10it
Date : March 29 2020, 07:55 AM
will help you I'm going to assume you are using Python 2 here, but write the code to work on both Python 2 and Python 3.
You have UTF-16 data, not UTF-8. You can read that as binary data and split on the four NUL bytes with the str.split() method:
code :
file.read(size).split(b'\x00' * 4)[index]
file.read(size).decode('utf-16-le').split(u'\x00' * 2)[index]
def str_extract(file, start, size, delimiter = None, index = None):
   if (delimiter is not None and index is not None):
       delimiter = delimiter.decode('utf-16-le')  # or pass in Unicode
       return file.read(size).decode('utf-16-le').split(delimiter)[index]
       return file.read(size).decode('utf-16-le')

with open('filename', 'rb') as fobj:
    result = str_extract(fobj, 0, 0x20, b'\x00' * 4, 0)
import io

with io.open('filename', 'r', encoding='utf16') as fobj:
    # ....
>>> from io import BytesIO
>>> data = b'H\x00e\x00l\x00l\x00o\x00 \x00W\x00o\x00r\x00l\x00d\x00\x00\x00\x00\x001\x002\x003\x00'
>>> fobj = BytesIO(data)
>>> str_extract(fobj, 0, 0x20, '\x00' * 4, 0)
u'Hello World'
>>> str_extract(fobj, 0, 0x20, '\x00' * 4, 1)
Regex search with variable in Python 2.7 returns bytes instead of decoded text

Regex search with variable in Python 2.7 returns bytes instead of decoded text

By : Sagar Modi
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Python 2.x only
Your find is probably not working because you're mixing strs and Unicodes strs, or strs containing different encodings. If you don't know what the difference between Unicode str and str, see: https://stackoverflow.com/a/35444608/1554386
code :
with io.open('C:\....txt', 'r', encoding='utf-8') as my_file:
    my_file_test = my_file.read()
# finds lines beginning with i, ending in .
regex = u'^{i}*?\.$'.format(i=i)
wantedtext = re.findall(regex, my_file_test, re.M)
with io.open('C:\....txt', 'r', encoding='utf-8') as f:
    wordslist = f.read().splitlines() 
convert hexadecimal to bytes in python

convert hexadecimal to bytes in python

By : Cornita Pinchinat
Date : March 29 2020, 07:55 AM
This might help you I want to display the number , Maybe not the most flexible way, but:
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