python如何将特殊字符串u'\\ x08'u'\\ x09'u'\\ x03'转换为8 9 3

Question

use crypto encrypt string in Go.

PKCS7 Padding for ciphertext, because ciphertext must be a multiple of the block size (16).

exa:   ciphertext is : 123456789abcde. (len:14) 
       pandding:  []byte("123456789abcdef") + []byte(2) + []byte(2)

Python decode:

list: [u'1', u'2', u'3', ...,u'd', u'e', u'\x02', u'\x02']

now, can not get u'\\x02' to number 2.

Answer 1

I'm sure this is not the best way to do this in python, but I tend to think in C, so here goes. It is probably pretty bad according to most people, but it gets the job done.

c = u'\x02'
byte = bytearray(c, 'utf-8')[0]
print(chr(ord('0') + byte))
=> 2

Answer 2

Inspired by andars

c = ord(u'\x02') + ord('0')
print chr(c)