Thanks in advance. I have an issue where merging some work produces two guids for a single subject in my triplestore. I need a way to isolate all of those instances so that I can normalize those guids. I'm new to SPARQL. I've tried returning these by COUNT and FILTER, but have so far failed. Here's what we're looking at.
<http://example.com/#bar> <http://www.w3.org/1999/02/22-rdf-syntax-ns#type>
<http://example.com/#choo> .
<http://example.com/#bar> <http://www.w3.org/1999/02/22-rdf-syntax-ns#type>
<http://example.com/#terms> .
<http://example.com/#bar> <http://www.w3.org/2008/05/skos-xl#prefLabel>
<http://example.com/#bar/bar_en> .
<http://example.com/#bar> <http://www.foo.com/#guid> "17bda3bb-7db9-4afa-
bb95-29da6464f137" .
<http://example.com/#bar> <http://www.foo.com/#guid> "1fa33bad-a98d-4a7e-
8679-d8777e690c0c" .
Sorry, that wrapped text makes this harder to read. However, you can see here that I have a subject, http://example.com/#bar which has two guids. I have many cases like this in the same graph and I need a way to identify them all. Thanks again.
As alluded to in @AKSW's comment you should be able to use HAVING to filter on aggregations combined with grouping by the subject to achieve this:
SELECT ?subject (COUNT(?guid) AS ?numGuids)
WHERE
{
?subject <http://www.foo.com/#guid> ?guid .
}
GROUP BY ?subject
HAVING (?numGuids > 1)