SPARQL Wikidata Query to retrieve url for multiple languages - sparql

Question

I wrote this Sparql query to retrieve item labels in multiple language
SELECT ?item ?en ?es ?it WHERE {
{ ?item wdt:P31 wd:Q6256. }
UNION
{ ?item wdt:P31 wd:Q1250464. }
UNION
{ ?item wdt:P31 wd:Q3624078. }
UNION
{ ?item wdt:P31 wd:Q619610. }
UNION
{ ?item wdt:P31 wd:Q179164. }
UNION
{ ?item wdt:P31 wd:Q7270. }
?item rdfs:label ?en filter (lang(?en) = "en").
?item rdfs:label ?es filter (lang(?es) = "es").
?item rdfs:label ?it filter (lang(?it) = "it").
} LIMIT 2
that gives
{
"item": {
"type": "uri",
"value": "http://www.wikidata.org/entity/Q43"
},
"en": {
"xml:lang": "en",
"type": "literal",
"value": "Turkey"
},
"es": {
"xml:lang": "es",
"type": "literal",
"value": "Turquía"
},
"it": {
"xml:lang": "it",
"type": "literal",
"value": "Turchia"
}
}
I would like now to retrieve item url for each language. An approach could be the following
SELECT ?cid ?country ?article_en ?article_de WHERE {
?cid wdt:P31 wd:Q3624078 .
OPTIONAL {
?cid rdfs:label ?country filter (lang(?country) = "en") .
}
OPTIONAL {
?cid rdfs:label ?country filter (lang(?country) = "de") .
}
OPTIONAL {
?article_en schema:about ?cid .
?article_en schema:inLanguage "en" .
FILTER (SUBSTR(str(?article_en), 1, 25) = "https://en.wikipedia.org/")
}
OPTIONAL {
?article_de schema:about ?cid .
?article_de schema:inLanguage "de" .
FILTER (SUBSTR(str(?article_de), 1, 25) = "https://de.wikipedia.org/")
}
}
that gives
{
"cid": {
"type": "uri",
"value": "http://www.wikidata.org/entity/Q236"
},
"article_de": {
"type": "uri",
"value": "https://de.wikipedia.org/wiki/Montenegro"
},
"article_en": {
"type": "uri",
"value": "https://en.wikipedia.org/wiki/Montenegro"
},
"country": {
"xml:lang": "en",
"type": "literal",
"value": "Montenegro"
}
}
Try it here.
This will require to have a separate property article_* for each language.
Putting all together:
SELECT ?item ?en ?es ?url_en ?url_es WHERE {
{ ?item wdt:P31 wd:Q43229. }
UNION { ?item wdt:P31 wd:Q4830453. }
OPTIONAL { ?item rdfs:label ?en filter (lang(?en) = "en"). }
OPTIONAL { ?item rdfs:label ?es filter (lang(?es) = "es"). }
OPTIONAL {
?url_en schema:about ?item .
?url_en schema:inLanguage "en" .
FILTER (SUBSTR(str(?url_en), 1, 25) = "https://en.wikipedia.org/")
}
OPTIONAL {
?url_es schema:about ?item .
?url_es schema:inLanguage "es" .
FILTER (SUBSTR(str(?url_es), 1, 25) = "https://es.wikipedia.org/")
}
} LIMIT 10
Try it here.
that gives
{
"item": {
"type": "uri",
"value": "http://www.wikidata.org/entity/Q130178"
},
"en": {
"xml:lang": "en",
"type": "literal",
"value": "Arcadie"
},
"es": {
"xml:lang": "es",
"type": "literal",
"value": "Arcadie"
},
"url_en": {
"type": "uri",
"value": "https://en.wikipedia.org/wiki/Arcadie"
},
"url_es": {
"type": "uri",
"value": "https://es.wikipedia.org/wiki/Arcadie_(grupo)"
}
}
Is there a way to group the url and the label property result like
{
"en": {
"url": {
"xml:lang": "es",
"type": "literal",
"value": "Arcadie"
},
"label": {
"type": "uri",
"value": "https://es.wikipedia.org/wiki/Arcadie_(grupo)"
}
}
}