Suppose I have a uri http://dbpedia.org/page/Manmohan_Singh now he has a list of years in his tag dbpprop:years.
When I write a query like
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX dbpedia: <http://dbpedia.org/resource/>PREFIX dcterms: <http://purl.org/dc/terms/>
PREFIX dbpedia-owl: <http://dbpedia.org/ontology/>PREFIX category: <http://dbpedia.org/resource/Category:>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>PREFIX foaf: <http://xmlns.com/foaf/0.1/>PREFIX dbpprop: <http://dbpedia.org/property/>
PREFIX dbprop: <http://dbpedia.org/property/>PREFIX grs: <http://www.georss.org/georss/>
PREFIX category: <http://dbpedia.org/resource/Category:>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX dbpprop: <http://dbpedia.org/property/>
PREFIX foaf: <http://xmlns.com/foaf/0.1/>
SELECT DISTINCT ?x ?name ?abs ?birthDate ?birthplace ?year ?party ?office ?wiki WHERE {
?x owl:sameAs? dbpedia:Manmohan_Singh.
?x dbpprop:name ?name.
?x dbpedia-owl:birthDate ?birthDate.
?x dbpedia-owl:birthPlace ?birthplace.
?x dbpprop:years ?year.
?x dbpprop:party ?party.
?x dbpedia-owl:office ?office.
?x foaf:isPrimaryTopicOf ?wiki.
?x rdfs:comment ?abs.
FILTER(lang(?abs) = 'en')
}
I get result of each year in different row .. and hence repeating data for other collumns. Is there a way I can get it as a list in just one collumn like all the years in one collumn comma separated or smthng like that?
Similary for the prop dbpedia-owl:office
This is similar to Aggregating results from SPARQL query, but the problem is actually a bit more complex, because there are multiple variables that have more than one result. ?name
, ?office
, and ?birthPlace
have the same issue.
You can work around this using group_concat
, but you'll need to use distinct
as well, to keep from getting, e.g., the same ?year
repeated multiple times in your concatenated string. group by
reduces the number of rows that you have in a solution, but in each of those rows, you have a set of values for the variables that you didn't group by. E.g., since ?year
isn't in the group by
, you have a set of values for ?year
, and you have to do something with them. You could, e.g., select (sample(?year) as ?aYear)
to grab just one from the set, or you could do as we've done here, and select (group_concat(distinct ?year;separator=", ") as ?years)
to concatenate the distinct values into a string.
You'll want a query like the following, which produces one row:
SELECT ?x
(group_concat(distinct ?name;separator="; ") as ?names)
?abs
?birthDate
(group_concat(distinct ?birthplace;separator=", ") as ?birthPlaces)
(group_concat(distinct ?year;separator=", ") as ?years)
?party
(group_concat(distinct ?office;separator=", ") as ?offices)
?wiki
WHERE {
?x owl:sameAs? dbpedia:Manmohan_Singh.
?x dbpprop:name ?name.
?x dbpedia-owl:birthDate ?birthDate.
?x dbpedia-owl:birthPlace ?birthplace.
?x dbpprop:years ?year.
?x dbpprop:party ?party.
?x dbpedia-owl:office ?office.
?x foaf:isPrimaryTopicOf ?wiki.
?x rdfs:comment ?abs.
FILTER(langMatches(lang(?abs),"en"))
}
group by ?x ?abs ?birthDate ?party ?wiki
SPARQL results
Look at GROUP_CONCAT
but it may be easier to retrieve the data in multiple rows (you can sort to put repeats adjacent to each other) and process in code.
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