英文:
SPARQL query to get an object that is shared (connected to) by two subjects
问题
以下是翻译好的内容：
我有以下图表
```sparql
@prefix : <http://example.org/> .
:JupiterAtmsphere :isComposedOf :Nitrogen.
:JupiterAtmsphere :isComposedOf :Helium.
:EarthAtmosphere :isComposedOf :Nitrogen.
:comet_1 :hasDiscoverer :JohnSmith.
:comet_5 :hasDiscoverer :JohnSmith.
:comet_3 :hasDiscoverer :JaneLee.
我想返回所有由多个主体共享的谓词和对象。该查询试图回答以下问题：哪些“主体”共享了“对象”？我希望查询返回:Nitrogen作为由:JupiterAtmsphere和:EarthAtmsphere共享的对象，以及:comet_1和:comet_5共享的对象:JohnSmith。然而，挑战在于我希望查询是通用的（没有关于图表的主体、谓词或对象的先验知识）。换句话说，我正在设计一个通用的查询，可以应用于不同的本体，而不考虑它们的词汇。
我尝试了以下查询，但没有产生期望的结果：
SELECT ?class1 ?class2 ?p1 ?o
WHERE {
?class1 ?p1 ?o.
?class2 ?p2 ?o.
FILTER (?p1 = ?p2)
<details>
<summary>英文:</summary>
I have the following graph
@prefix : <http://example.org/> .
:JupiterAtmsphere :isComposedOf :Nitrogen.
:JupiterAtmsphere :isComposedOf :Helium.
:EarthAtmosphere :isComposedOf :Nitrogen.
:comet_1 :hasDiscoverer :JohnSmith.
:comet_5 :hasDiscoverer :JohnSmith.
:comet_3 :hasDiscoverer :JaneLee.
I want to return all predicates and objects that are shared by a number of subjects. The query is trying to answer the following question: Which 'subjects' share 'object' in common?. I'd like the query to return :Nitrogen as an object that is shared by both :JupiterAtmspher and :EarthAtmspher, and :JohnSmith as an object that is shared by :comet_1 and :comet_5
However, the challenge is that I want the query to be generic (without prior knowledge of the subjects, predicates or objects of the graph). In other words, I'm desinging a query that is generic and can be applied to different ontologies regardless of their vocabulary.
I tried the following query which did not yeild the desired result:
SELECT ?class1 ?class2 ?p1 ?o
WHERE {
?class1 ?p1 ?o.
?class2 ?p2 ?o.
FILTER (?p1 = ?p2)
</details>
# 答案1
**得分**: 0
你必须声明主语必须不同。您可以在`FILTER`中使用`!=`运算符来实现：
```sparql
SELECT ?s1 ?s2 ?p1 ?o
WHERE {
?s1 ?p1 ?o .
?s2 ?p2 ?o .
FILTER (?s1 != ?s2) .
} ORDER BY ?o ?p1
如果谓词需要相同，将?p1和?p2替换为?p。
（注意：此查询找到至少两个主语共享的对象。）
使用GROUP_CONCAT，您可以获得每个对象+谓词的一个结果解决方案（例如，带有逗号分隔的主语列表）。
英文:
You have to state that the subjects need to be different. You can do this with the != operator in a FILTER:
SELECT ?s1 ?s2 ?p1 ?o
WHERE {
?s1 ?p1 ?o .
?s2 ?p2 ?o .
FILTER (?s1 != ?s2) .
} ORDER BY ?o ?p1
If the predicates need to be the same, replace ?p1 and ?p2 with ?p.
(Note: This query finds objects that are shared by at least two subjects.)
With GROUP_CONCAT, you could get one result solution per object+predicate (e.g., with a comma-separated list of subjects).
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