Hello Jindřich,
how do you apply both FILTERs? Obviously the filter on the length needs
to be evaluated first, and I'm wondering how this would be evaluated if
somebody uses FILTER(cond1 && cond2) as && should be commutative.
Does it work if you use a sub-SELECT to filter by length first?
Kind regards,
Lorenz
> Unfortunately, using STRDT(?literal, xsd:string) doesn't make
> difference and the error is still thrown. ?literal is already a string.
>
> - Jindřich
>
> --
> Jindřich Mynarz
> http://mynarz.net/#jindrich
>
> On Tue, Nov 1, 2016 at 2:49 PM, Nikolaos Beredimas <bere...@gmail.com
> <mailto:bere...@gmail.com>> wrote:
>
> I've had similar problems lately working with strings.
> SPARQL snippets unctions that worked on other products and
> produced this kind of errors in Virtuoso.
> Try using:
> STRDT(?literal, xsd:string)
> as input to the substr() function,
> and it should work.
>
> On Tue, Nov 1, 2016 at 3:22 PM, Jindřich Mynarz
> <mynarzjindr...@gmail.com <mailto:mynarzjindr...@gmail.com>> wrote:
>
> Hi,
>
> I have a query that first filters literals of given length:
>
> FILTER (STRLEN(?literal) = 8)
>
> And then it extracts a substring via SUBSTR():
>
> SUBSTR(?literal, 8, 1)
>
> However, Virtuoso throws Virtuoso 22011 Error SR026: SPARQL
> substr: Bad string subrange: from=8, len=1.
> How is that possible?
> I'm using the develop version of Virtuoso
>
> (https://github.com/openlink/virtuoso-opensource/commit/377438ac6beefa9373cf0500c7edf4077f687d0c
>
> <https://github.com/openlink/virtuoso-opensource/commit/377438ac6beefa9373cf0500c7edf4077f687d0c>).
> - Jindřich
> --
> Jindřich Mynarz
> http://mynarz.net/#jindrich
>
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--
Lorenz Bühmann
AKSW group, University of Leipzig
Group: http://aksw.org - semantic web research center
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