Question

有没有办法过滤像

这样的查询

select ?x ?y  where {?x <http://relationship.com/wasRevisionOf>+ ?y }";

对于提供的数据集具有以下输出：

http://article.com/2-3 http://article.com/2-2
http://article.com/2-3 http://article.com/2-1
http://article.com/2-4 http://article.com/2-3
http://article.com/2-4 http://article.com/2-2
http://article.com/2-4 http://article.com/2-1
http://article.com/2-2 http://article.com/2-1
http://article.com/1-3 http://article.com/1-2
http://article.com/1-3 http://article.com/1-1
http://article.com/1-2 http://article.com/1-1

我们如何过滤查询，以便我们删除所有结果，其中？x值等于另一个结果中的？y值。通过这个，我们将得到

http://article.com/2-4 http://article.com/2-3
http://article.com/2-4 http://article.com/2-2
http://article.com/2-4 http://article.com/2-1

因为所有其他结果的？x值在另一个结果中作为？y值出现。

这是数据集：

<http://article.com/1-3> <http://relationship.com/wasGeneratedBy> <http://edit.com/comment1-2> .
<http://article.com/1-3> <http://relationship.com/wasRevisionOf> <http://article.com/1-2> .
<http://edit.com/comment1-2> <http://relationship.com/used> <http://article.com/1-2> .
<http://edit.com/comment1-2> <http://relationship.com/wasAssociatedWith> <http://editor.com/user1-1> .

<http://article.com/1-2> <http://relationship.com/wasGeneratedBy> <http://edit.com/comment1-1> .
<http://article.com/1-2> <http://relationship.com/wasRevisionOf> <http://article.com/1-1> .
<http://edit.com/comment1-1> <http://relationship.com/used> <http://article.com/1-1> .
<http://edit.com/comment1-1> <http://relationship.com/wasAssociatedWith> <http://editor.com/user1-1> .

<http://article.com/2-4> <http://relationship.com/wasGeneratedBy> <http://edit.com/comment2-3> .
<http://article.com/2-4> <http://relationship.com/wasRevisionOf> <http://article.com/2-3> .
<http://edit.com/comment2-3> <http://relationship.com/used> <http://article.com/2-3> .
<http://edit.com/comment2-3> <http://relationship.com/wasAssociatedWith> <http://editor.com/user2-3> .

<http://article.com/2-3> <http://relationship.com/wasGeneratedBy> <http://edit.com/comment2-2> .
<http://article.com/2-3> <http://relationship.com/wasRevisionOf> <http://article.com/2-2> .
<http://edit.com/comment2-2> <http://relationship.com/used> <http://article.com/2-2> .
<http://edit.com/comment2-2> <http://relationship.com/wasAssociatedWith> <http://editor.com/user2-2> .

<http://article.com/2-2> <http://relationship.com/wasGeneratedBy> <http://edit.com/comment2-1> .
<http://article.com/2-2> <http://relationship.com/wasRevisionOf> <http://article.com/2-1> .
<http://edit.com/comment2-1> <http://relationship.com/used> <http://article.com/2-1> .
<http://edit.com/comment2-1> <http://relationship.com/wasAssociatedWith> <http://editor.com/user2-1> .

Answer 1

select ?x ?y  where {?x <http://relationship.com/wasRevisionOf>+ ?y }
我们如何过滤查询，以便我们删除所有结果 a？x值等于另一个结果中的？y值。

如果一行中的？x值是另一行中的？y值，则表示？x是 wasRevisionOf 属性中某个三元组的对象。你可以简单地过滤掉那些：

select ?x ?y  where {
  ?x <http://relationship.com/wasRevisionOf>+ ?y
  filter not exists {
    ?something <http://relationship.com/wasRevisionOf> ?x
  }
}

这确保了？x的每个值都是链的“开头”。

在SPARQL 1.1中过滤属性路径

1 个答案: