Question

根据这些数据，每个人可以选择性地拥有＆＃34; smart＆＃34;谓词，每个部门可能有零个或多个人，我需要找到只包含聪明人的部门。结果应该只包括部门1和2.理想情况下，结果还应包括＆＃34; smart＆＃34;每个部门的对象。谢谢！

composer update

Answer 1

像双重否定之类的东西可能会起作用：

SELECT DISTINCT ?dept WHERE {
  ?dept p:has ?person .
  FILTER NOT EXISTS {
  ?dept p:has ?person1 .
    FILTER NOT EXISTS {
     ?person1 p:smart ?smartVal
    }
  }
}

结果：

+---------------+
|     dept      |
+---------------+
|  department:1 |
|  department:2 |
+---------------+

使用值：

SELECT ?dept (GROUP_CONCAT(DISTINCT ?smart;separator=";") as ?smartValues) WHERE {
  ?dept p:has ?person .
  ?person p:smart ?smart
  FILTER NOT EXISTS {
  ?dept p:has ?person1 .
    FILTER NOT EXISTS {
     ?person1 p:smart ?smartVal
    }
  }
}
GROUP BY ?dept

结果：

+---------------+-------------+
|     dept      | smartValues |
+---------------+-------------+
|  department:1 | maybe;yes   |
|  department:2 | maybe       |
+---------------+-------------+

Answer 2

我有一种感觉我之前已经回答了类似的事情，但无论如何有一个相当不错的方法来做到这一点：

select ?dept 
   (count(?person) as ?pc) (count(?smart) as ?sc)
   (group_concat(?smart; separator=',') as ?smarts)
{
    ?dept p:has ?person .
    optional { ?person p:smart ?smart }
}
group by ?dept
having (?pc = ?sc)

即：找到部门，人员和（可用的）智能价值。对于每个部门，找到人数与智能值数量匹配的部门。

-------------------------------------------------------------
| dept                              | pc | sc | smarts      |
=============================================================
| <http://example.com/department#2> | 1  | 1  | "maybe"     |
| <http://example.com/department#1> | 2  | 2  | "yes,maybe" |
-------------------------------------------------------------

当您想要获得每个对象的结果时，匹配某些条件group by / having通常是最干净的答案（因为您可以将匹配与过滤分开）。

SPARQL：查找所有符合条件的子对象的对象

2 个答案: