Sparql，如何合並不同的結果

Question

我試圖創建一個SPARQL查詢，以查找Jim認識的所有人員，然后查找Jim的認識者認識的人員，然后找到類似鏈的東西。

例如，我有：

Jim knows Clare and Antoine    
Clare knows Jim and David
Antoine knows David and Clare
David knows Clare

因此結果如下：

result1: Clare, Antoine
result2: Jim, David, David, Clare
result3: Clare, David, Clare, Clare, Jim, David

我或多或少地做了類似樹的東西。

我想要的是將result1 ， result2和result3合並到result4 。 因此， result4將是：

result4: clare, antoine, jim, david, david, clare, clare, david, clare, clare, jim, david.

然后使用DISTINCT刪除重復項。 請問我該如何實現？

SELECT  ?Result1 ?Result2 ?Result3
WHERE{
    {
        base:Knows  dc:Names        _:BN1 .
        _:BN1       dc:FName        "Jim";
                dc:KnownFName           ?Result1 .  
    }
    .
    {
        base:Knows  dc:Names        _:BN2 .
        _:BN2       dc:FName        ?Result1;
                dc:KnownFName           ?Result2 .
    }
    .   
    {
        base:Knows  dc:Names        _:BN3 .
        _:BN3       dc:FName        ?Result2;
                dc:KnownFName           ?Result3 .
    }
}

Answer 1

假設您的示例表示如下：

<http://example.com/person/1> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/2> .
<http://example.com/person/1> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/3> .
<http://example.com/person/2> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/1> .
<http://example.com/person/2> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/4> .
<http://example.com/person/3> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/4> .
<http://example.com/person/3> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/2> .
<http://example.com/person/4> <http://xmlns.com/foaf/0.1/knows> <http://example.com/person/2> .

<http://example.com/person/1> <http://www.w3.org/2000/01/rdf-schema#label> "Jim" .
<http://example.com/person/2> <http://www.w3.org/2000/01/rdf-schema#label> "Clare" .
<http://example.com/person/3> <http://www.w3.org/2000/01/rdf-schema#label> "Antoine" .
<http://example.com/person/4> <http://www.w3.org/2000/01/rdf-schema#label> "David" .

然后，您可以使用SPARQL的UNION功能，通過合並三個獨立查詢的結果來實現所需的功能。 可以使用SPARQL屬性路徑更簡潔地表示它：

PREFIX foaf: <http://xmlns.com/foaf/0.1/>

SELECT DISTINCT * WHERE {
{ SELECT * WHERE {
<http://example.com/person/1> foaf:knows/rdfs:label ?knowsName .
} }
UNION
{ SELECT * WHERE {
<http://example.com/person/1> foaf:knows/foaf:knows/rdfs:label ?knowsName .
} }
UNION
{ SELECT * WHERE {
<http://example.com/person/1> foaf:knows/foaf:knows/foaf:knows/rdfs:label ?knowsName .
} }
}

您還可以通過單個屬性路徑表達式來獲得knows謂詞的完全可傳遞閉包：

PREFIX foaf: <http://xmlns.com/foaf/0.1/>

SELECT * WHERE {
<http://example.com/person/1> foaf:knows+/rdfs:label ?knowsName .
} 

...如果您的Triplestore支持SPARQL 1.1。 否則，您將不得不使用推理或重復查詢來獲得完全關閉。