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Subject: Bug in setRawPath method of URI class
Date: Fri, 24 Jan 2003 10:50:04 +0100
Message-ID: <FF2B76E6B7DB08469C3E8E96E3F3BC3F01EAA4B4@skywalker.newknow.com>
Thread-Topic: Bug in setRawPath method of URI class
Thread-Index: AcLDjfiqKquxABF8QzmdE4QMApkIQw==
From: =?iso-8859-1?Q?Armando_Ant=F3n?= <armando.anton@newknow.com>
To: "Commons HttpClient Project (E-mail)"
 <commons-httpclient-dev@jakarta.apache.org>

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Hi!

i think there is a bug in the setRawPath method when working with =
relative paths that does not start with the '/' character

i will write first the code and after that i will try to explain why =
there is a bug :)

class URI (lines 2692-2701)
        } else if (_is_rel_path) {
            int at =3D indexFirstOf(escapedPath, '/');
            if (at =3D=3D 0)
                throw new URIException(URIException.PARSING, "incorrect =
path");
            if (at > 0 && !validate(escapedPath, 0, at, rel_segment) ||
                    at < 0 && !validate(escapedPath, 0, -1, abs_path))
                throw new URIException(URIException.ESCAPING,
                        "escaped relative path not valid");
            _path =3D escapedPath;
        }

when the uri is relative, first it checks if the escapedPath starts with =
the '/' character (if true then the path is not relative and an =
exception is thrown)
after that, there is a double check:
1. there is a '/' character and the escapedpath(from the first character =
to the first occurrence of the '/' character) is valid (using the =
rel_segment BitSet)
2. there is not a '/' character and the escapedpath (all the characters) =
is valid (using the abs_path BitSet)

the problem is in the first check because the substring validated ends =
with the '/' character and the rel_segment BitSet does not allow that =
character
so an exception is thrown

i really does not understand that double check and i have changed it to:

        } else if (_is_rel_path) {
            int at =3D indexFirstOf(escapedPath, '/');
            if (at =3D=3D 0)
                throw new URIException(URIException.PARSING, "incorrect =
path");
            if (!validate(escapedPath, 0, -1, abs_path))
                throw new URIException(URIException.ESCAPING,
                        "escaped relative path not valid");
            _path =3D escapedPath;
        } =20

To test this bug you can execute this simple class and an exception will =
occur

import org.apache.commons.httpclient.URI;

public class TestSetRawPath {

    public static void main(String[] args) throws Exception {
        char[] _uri =3D "dir/doc.html".toCharArray();
        URI uri =3D new URI(_uri);
    }
}

Hope it helps,

Thank you for your great effort,

Armando
 <<URI_v1.22.diff>>=20

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