SPARQL查询以获取节点的所有父节点

Question

我的mySQL数据库中有表格,如:

+-------+------------+-------------+-----------+
|  ID   |  subject   |  Predicate  |  object   |
+-------+------------+-------------+-----------+
|  1    | ATM        |  subClassof | Network   |
+-------+------------+-------------+-----------+
|  2    | ARPANET    |  subClassof | Network   |
+-------+------------+-------------+-----------+
|  3    | Network    |  subClassof | Main      |
+-------+------------+-------------+-----------+
|  5    | Software   |  subclassof | Main      |
+-------+------------+-------------+-----------+
|  7    | Linux      |  subClassof | Software  |
+-------+------------+-------------+-----------+
|  8    | Windows    |  subClassof | Software  |
+-------+------------+-------------+-----------+
|  12   | XP         |  subClassof | Windows   |
+-------+------------+-------------+-----------+
|  13   | Win7       |  subClassof | Windows   |
+-------+------------+-------------+-----------+
|  14   | Win8       |  subClassof | Windows   |
+-------+------------+-------------+-----------+

对于Predicate subClassof,它将具有如下树视图:

Main
   |__ Network
   |         |__ ATM
   |         |__ ARPANET
   |
   |__ Software
              |__ Linux
              |__ Windows
                        |__ XP
                        |__ Win7
                        |__ Win8

我想创建一个表单,可以选择起始节点并获取所有父节点.例如,通过选择Win7我想得到:

main, Software, Windows,Win7

---

Step2:有没有办法用这样的简单文本打印这个节点:

Main
   |__ Software
              |__ Windows
                        |__ Win7

Answer 1

您的数据可以在RDF中表示为data.n3:

@prefix : <http://example.org/> .
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .

:Network rdfs:subClassOf :Main .

:ATM rdfs:subClassOf :Network .
:ARPANET rdfs:subClassOf :Network .

:Software rdfs:subClassOf :Main .

:Linux rdfs:subClassOf :Software .
:Windows rdfs:subClassOf :Software .

:XP rdfs:subClassOf :Windows .
:Win7 rdfs:subClassOf :Windows .
:Win8 rdfs:subClassOf :Windows .

从这里开始,您只需要一个SPARQL查询,该查询通过rdfs:subClassOf属性的路径(包括空路径)查找连接到特定类的所有内容.

prefix : <http://example.org/>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>

select ?superclass where { 
  :Win7 rdfs:subClassOf* ?superclass
}

--------------
| superclass |
==============
| :Win7      |
| :Windows   |
| :Software  |
| :Main      |
--------------

该查询中的结果不一定按它们在路径中的位置排序(尽管在这种情况下它们恰好是).如果你确实需要它们,你可以这样做(这是基于这个关于计算RDF列表中元素位置的答案):

prefix : <http://example.org/>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>

select ?class where { 
  :Win7 rdfs:subClassOf* ?mid .
  ?mid rdfs:subClassOf* ?class .
}
group by ?class
order by count(?mid)

此找到的每个祖先?class的:Win7以及每个?mid中间祖先.对于祖先?class,距离计算为(count(?mid))之间的中间关系的数量.它根据距离对结果进行排序,:Win7最接近的祖先:Windows也是如此,依此类推.

你甚至可以做一些你想要的花哨格式:

prefix : <http://example.org/>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>

select (group_concat( ?name ; separator="--" )  as ?path) where {
  {
    select ?name where { 
      :Win7 rdfs:subClassOf* ?mid .
      ?mid rdfs:subClassOf* ?class .
      bind( strAfter( str(?class), "http://example.org/") as ?name )
    }
    group by ?class ?name
    order by count(?mid)
  }
}

-----------------------------------
| path                            |
===================================
| "Win7--Windows--Software--Main" |
-----------------------------------

有可能做一些更高级的字符串处理并获得多行字符串.你可能会看一下这个答案的后半部分,其中有一些花哨的格式,可以很好地对齐一个想法的矩阵.