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Problems with Regex when using SPARQL REPLACE function with certain characters #415

qtips opened this Issue May 26, 2015 · 1 comment


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qtips commented May 26, 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 ofÆØÅæøå

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


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jindrichmynarz commented Nov 20, 2016

I stumbled upon a similar issue. When I use a non-ASCII character in the replacement, it ends up broken:

SELECT (REPLACE(".", "\\.", "á") AS ?s)
# Result: á
# Expected: á 

However, when STR() is applied to the replacement, what should be in theory a no-op delivers the correct result:

SELECT (REPLACE(".", "\\.", STR("á")) AS ?s)
# Result: á
# Expected: á 

Unfortunately, this work-around doesn't work for all non-ASCII characters:

SELECT (REPLACE(".", "\\.", STR("š")) AS ?s)
# Result: ?
# Expected: š

This seems to be a general problem of STR(), which I've filed as #609.

Tested using Virtuoso version 07.20.3217.

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