Virtuoso SPARQL Replace Function Bug and Workaround

05 Jul 2015

Running the following SPARQL for replacing %c3%85 with the letter Å runs as expected

select REPLACE("%c3%85-XYZ-%20%28-DEF-%29","%C3%85", "Å", 'i')
where {}

#Result: Å-XYZ-%20%28-DEF-%29

However, when using nested REPLACE statements with an outer replace having a regex with ., the replace function “jumps” back one character where the match is found :

select REPLACE(
             REPLACE("%c3%85-XYZ-%20%28-DEF-%29","%C3%85", "Å", 'i'),
          "%..(%..)*", "?", 'i')

where {}
# Result: Å-XYZ?8-DEF?9
# Expected: Å-XYZ-?-DEF-?

This only happens for some replace characters including all of ÆØÅæøå. Workaround for this is to run a CONCAT before the second REPLACE, which seems to “reset” the string before sending it to next REPLACE:

select REPLACE(

CONCAT(REPLACE("%c3%85-XYZ-%20%28-DEF-%29","%C3%85", "Å", 'i'),""),

"%..(%..)*", "?", 'i')

where {}

# Result: Å-XYZ-?-DEF-?

This was tested using Virtuoso version 07.20.3212 on Linux (x86_64-unknown-linux-gnu), Single Server Edition with Virtuoso SPARQL Query Editor

I also made an issue on Github at Openlink/Virtuoso in case they fix this :)