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Home/ Questions/Q 6852995
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T01:27:30+00:00 2026-05-27T01:27:30+00:00

I’m a beginner to Python trying to decode this javascript sequence. I’m not only

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I’m a beginner to Python trying to decode this javascript sequence. I’m not only a beginner in Python, I understand javascript even less 🙂 I know i could put it into an online decoder such as this: http://coderstoolbox.net/string/ but I want to deal with it myself – more for practice than anything else… Im using Beautiful Soup to get the data, so I have its functions available for decoding.

If anyone can point me to equivalent functions in python I would appreciate it.

function encode(str){
 var t=''; var s=unescape(str);
 var x=Math.round(Math.sqrt(49));
 for(var i=0;i<s.length;i++) t+=
  String.fromCharCode(s.charCodeAt(i)^(i%2?x:0));
 print(t);
}

This is my understanding of it so far:

  1. i think I can use ‘HTML entities in BS to unescape..?
  2. the second one just seems to be a constant number ? square root of 49 rounded…
  3. sets up the loop
  4. this is the one i dont get. 1 i dont know what the fromCharCode function does. 2 not sure what the bit at the end is. looks like its getting a character code from i to the power something. i understand i is being modulo’d with 2 but what is the ‘?x:0’ bit ? – how would you replicate this in Python ?

thanks for reading !

EDIT: is there a python library that can just run this code ? I’ve done this before with bash and used rhino, but the alternatives in Python seem a bit scary for a beginner, eg spidermonkey, etc…

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  1. Editorial Team
    Editorial Team
    2026-05-27T01:27:30+00:00Added an answer on May 27, 2026 at 1:27 am

    1) the python equivalent to unescape is urllib.unquote() in python 2.x series and urllib.parse.unquote() in python 3.x series

    2) you guess the simplest way to do it is to do x = 7

    3) the simplest way to loop on string charters is to do for c in string: but to have the index you should do for i,c in enumerate(string):

    4) the string.charChodeAt(c) is the same than chr(c)

    finally I would duplicate the loop part as follow:

    result = []
    for i,c in enumerate(string):
        result.append(chr(ord(c)^(x if i%2 else 0)))
    print("".join(result))
    

    in fact using a temporary array to make the appends is more efficient than happening to strings as the strings don’t mutate

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