Hi.
I have a dataset, which has backslash character in some of the Subject URIs.
Whenever I try to use such URI in SPARQL Query, Virtuoso gives me this:
Virtuoso 37000 Error SP030: SPARQL compiler, line 0: Invalid character in
SPARQL expression at '\'
Do I need to escape backslash somehow? I tried to replace it with "%5C" and
"%5c" but that just gives me empty result set.
Example of URI:
<http://example.com/some/path/and\more>
------------------------------------------------------------------------------
Own the Future-Intel® Level Up Game Demo Contest 2013
Rise to greatness in Intel's independent game demo contest.
Compete for recognition, cash, and the chance to get your game
on Steam. $5K grand prize plus 10 genre and skill prizes.
Submit your demo by 6/6/13. http://p.sf.net/sfu/intel_levelupd2d
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