I have three hex strings:
hex1 = "e0"
hex2 = "aa"
hex3 = "b0"
string = "\\x"+hex1+"\\x"+hex2+"\\x"+hex3
print string
When I concatenate those three strings after appending "\x" to each of them I don't get its character representation.
I get output as \xe0\xaa\xb0
But when I define it in one line
string = "\xe0\xaa\xb0"
and print string I get correct output which is ર
What is wrong in my previous attempt?
Try
lst = [
chr(int(hex1, 16)),
chr(int(hex2, 16)),
chr(int(hex3, 16))
]
s = ''.join(lst) # '\xe0\xaa\xb0'
Your method won't work because the initial string "\\x" is interpreted as the string "\x" - and as you probably saw, creating the initial string with a single backslash ("\x") is invalid.
Python comes with a lot of useful libraries, and there is a Python library for what you want, binascii. binascii.unhexlify converts a hex sequence like '010203' into bytes '\x01\x02\x03':
>>> hex1 = "e0"
>>> hex2 = "aa"
>>> hex3 = "b0"
>>> s = hex1+hex2+hex3
>>> import binascii
>>> binascii.unhexlify(s)
'\xe0\xaa\xb0'
There is even a method for what you were originally trying to do:
>>> hex1 = "e0"
>>> hex2 = "aa"
>>> hex3 = "b0"
>>> s= "\\x"+hex1+"\\x"+hex2+"\\x"+hex3
>>> s.decode('string-escape')
'\xe0\xaa\xb0'
\\xand try to append.