Hi all,

I've been struggling with this for an hour or so, I'm trying to take a string such as

'Hell' to '\x48\x65\x6c\x6c'

For what I'm doing, I require the \x before the ascii hex

This is a dated solution from StackOverflow, that doesn't work on Python3

>>> mystring = "Hello World"
>>> print ''.join(r'\x{02:x}'.format(ord(c)) for c in mystring)
\x48\x65\x6c\x6c\x6f\x20\x57\x6f\x72\x6c\x64                

My new solution goes like this:

mystring = b'hello world'

#convert to hex
mystring = hexlify(mystring)
mystring = mystring.decode('ascii')
# hex, like this 68656c6c6f20776f726c64

#add in the \x in the right spots
mystring = '\\x'.join(mystring[i:i + 2] for i in range(-2, len(mystring), 2)) 
#looks like this at this point \x68\x65\x6c\x6c\x6f\x20\x77\x6f\x72\x6c\x64

#I need it back into the "b" format, so I encode it 
mystring=mystring.encode(ascii)
# This breaks everything, now it becomes:
# b'\\x68\\x65\\x6c\\x6c\\x6f\\x20\\x77\\x6f\\x72\\x6c\\x64'
# I specifically can't use the double backslashes.

Any ideas how to fix this or am I going about this all wrong? Thanks