Thanks for your quick reply :-)

But when I use an aggregates functions inside the INSERT clause Virtuoso
send me an error saying it was a syntax error.

I would like using this count result inside the INSERT not inside a SELECT.

Best.

Julien.


2013/9/19 Jim McCusker <mccus...@gmail.com>

> On virtuoso you need to use concat to count tuples of variables:
>
> select count(concat(?s,"|", ?o)) as ?indegree where {
>    ?s ?o <http://dbpedia.org/resource/Cat>.
> }
>
> will return 81 on http://dbpedia.org/sparql
>
>
> On Thu, Sep 19, 2013 at 11:51 AM, Julien Plu <
> julien....@redaction-developpez.com> wrote:
>
>>   Hi,
>>
>>  I have a problem with a SPARQL, I try to find a valid query for SPARQL
>> 1.0 through Virtuoso 06.01.3127 for doing a query like this :
>>
>> INSERT { GRAPH <http://my/graph> { ?s <http://my/ontology/indegree>
>> count(?x) }} where  {
>>    GRAPH <http://my/graph> {
>>       ?x ?p ?s .
>>       ?s a <http://my/ontology/Person>
>>   }
>> }
>>
>>  Most precisely I try to add a triple which represents the number of
>> triples which have for object a same subject for each person. For example
>> for this graph :
>>
>>  <http://Person/1> <http://my/ontology/knows> <http://Person/2> .
>>  <http://Person/3> <http://my/ontology/isBossOf> <http://Person/2> .
>> <http://Person/4> <http://my/ontology/isFatherOf> <http://Person/2> .
>>
>>  The result must be the triple : <http://Person/2> <
>> http://my/ontology/indegree> "3" .
>>
>>  Anyone knows how to do this easily just with a query ?
>>
>>  Thanks in advance.
>>
>>  Julien.
>>
>
>
>
> --
> Jim McCusker
>
> Data Scientist
>  5AM Solutions
> jmccus...@5amsolutions.com
> http://5amsolutions.com
>
> PhD Student
> Tetherless World Constellation
> Rensselaer Polytechnic Institute
> mcc...@cs.rpi.edu
> http://tw.rpi.edu
>
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