hex to string in javascript

Question

How can I convert: from: '\\x3c' to: '<';

I tried:

s=eval(s.replace("\\\\", ""));

does not work. How I do this? Thanks in advance!

score 12 · Accepted Answer · answered Aug 24 '11 at 15:30

12

Use String.fromCharCode instead of eval, and parseInt using base 16:

s=String.fromCharCode(parseInt(s.substr(2), 16));

answered Aug 24 '11 at 15:30

Digital Plane

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+1, a shorter alternative would be `String.fromCharCode(+("0"+s.slice(1)));` – Andy E Aug 24 '11 at 15:59
@Andy Even shorter: `String.fromCharCode("0"+s.slice(1));` – Digital Plane Aug 25 '11 at 06:55
@DigitalPlane: of course, `fromCharCode` casts to an int for you! doh! :-) – Andy E Aug 25 '11 at 08:35

score 1 · Answer 2 · edited May 23 '17 at 12:08

1

If you're using jQuery, try this: $('<div>').html('\x3c').text()

Else (taken from here)

function htmlDecode(input){
  var e = document.createElement('div');
  e.innerHTML = input;
  return e.childNodes.length === 0 ? "" : e.childNodes[0].nodeValue;
}

edited May 23 '17 at 12:08

Community

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answered Aug 24 '11 at 15:30

Mrchief

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score 0 · Answer 3 · answered Aug 24 '11 at 15:31

0

One way to do it, which will incur the wrath of people who believe that "eval" is unequivocally evil, is as follows:

var s = "\\x3c";
var s2 = eval('"' + s + '"'); // => "<"

answered Aug 24 '11 at 15:31

maerics

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I wouldn't say `eval` is *unequivocally* evil, but I would say that this isn't really an acceptable use of `eval` (since alternative options exist). – Andy E Aug 24 '11 at 16:08
@Andy E: ok, I'll take the bait =) I agree the `String.fromCharCode` is the right way but assuming we have validated the input (e.g. ensured it can only be an escaped character reference) then what's the potential harm in using `eval` here? To me it seems like one of the few times where it is ok. – maerics Aug 24 '11 at 16:47
I wasn't fishing ;-) But since you asked, `eval` is not only avoided for its security risks when passed unsanitised values, but also for its poor performance. Although `eval` itself is slow (since it invokes the js compiler), it also makes the code around it slow. The reason for this is, where a compiler would normally make optimizations as it interprets code, it cannot know the result of the eval'd expression and therefore cannot make such optimizations. There *are* uses for `eval`, but in the end it's the dev's decision to look at alternative solutions before taking the plunge. – Andy E Aug 25 '11 at 00:36

hex to string in javascript

3 Answers3