URLDecode.decode method not working as expected in Java

Question

I was trying to decode URL encoded post body and came across this problem.

I was using this method to decode (it decodes multiple encoded urls too) :

public static String decodeUrl(String url)
    {
        try {
            String prevURL="";
            String decodeURL=url;
            while(!prevURL.equals(decodeURL))
            {
                prevURL=decodeURL;
                decodeURL= URLDecoder.decode( decodeURL, "UTF-8" );
            }
            return decodeURL;
        } catch (UnsupportedEncodingException e) {
            return "Issue while decoding" +e.getMessage();
        }
    }

When the input url was "a%20%2B%20b%20%3D%3D%2013%25!" , the control somehow doens't show up after line decodeURL = when debugging . No exceptions are raised too.

The issue is that control doesn't go beyond the line"decodeURL" .

What might be causing the issue ? Please use debugger to hopefully mimic this problem.

what do you mean with "control" ?

k5_
– k5_

2018-08-08 19:01:24 +00:00
Commented Aug 8, 2018 at 19:01 — k5_
– k5_, Commented Aug 8, 2018 at 19:01

Giulio Piancastelli · Accepted Answer · 2018-08-08 19:12:37Z

2

Just tested it on Java 8u151. This throws an IllegalArgumentException on the second spin of the loop: "URLDecoder: Incomplete trailing escape (%) pattern". That's because after the first decoding you have "a + b == 13%!", and during the second decoding that % is supposed to introduce an encoding sequence but it does not. I think that's the expected behaviour, even if the standard library of other languages does not agree. Python 3.6 for example:

>>> from urllib.parse import unquote
>>> result = unquote('a%20%2B%20b%20%3D%3D%2013%25!')
>>> result
'a + b == 13%!'
>>> unquote(result)
'a + b == 13%!'

edited Aug 8, 2018 at 19:12

answered Aug 8, 2018 at 19:06

Giulio Piancastelli

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