SPARQL查询以获取节点的所有父节点 [英] SPARQL query to get all parent of a node

问题描述

我的mySQL数据库中有一些表，例如:

 +-------+------------+-------------+-----------+
|  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   |
+-------+------------+-------------+-----------+
 

对于谓词subClassof，它将具有以下树状视图:

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

我想创建一个表单，该表单可以选择起始节点并为此获取所有父节点.例如，通过选择Win7我想要得到:

main, Software, Windows, Win7

---

第2步:，可以使用以下简单文本来打印此节点:

 Main
   |__ Software
              |__ Windows
                        |__ Win7
 

解决方案

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事物.

 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其在路径中的位置排序(尽管在这种情况下它们恰好是).如果确实需要它们，则可以执行此操作(基于有关计算元素在RDF列表中的位置的答案):

 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)
 

这会找到:Win7的每个祖先?class以及每个?mid中间祖先.对于祖先?class，距离被计算为(count(?mid))之间的中间关系数.它根据该距离对结果进行排序，因此:Win7是最接近的祖先，:Windows在其后，依此类推.

您甚至可以像这样进行一些精美的格式化:

 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" |
-----------------------------------
 

可能可以进行一些更高级的字符串处理并获得多行字符串.您可能会看此答案的后半部分，其中有一些精美的格式，用于将思想很好地对齐在一起. >

I have tables in my mySQL database like:

+-------+------------+-------------+-----------+
|  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   |
+-------+------------+-------------+-----------+

For Predicate subClassof it will have a tree view like this:

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

I want to create a form which can select the start node and get all parent for that. For example by choosing Win7 i want to get:

main, Software, Windows,Win7

---

Step2: is there any way to print this nodes with a simple text like this:

Main
   |__ Software
              |__ Windows
                        |__ Win7

解决方案

Your data can be represented in RDF as 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 .

From here, you just want a SPARQL query that finds all the things connected to a particular class by a path (including the empty path) of rdfs:subClassOf properties.

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      |
--------------

The results in that query aren't necessarily ordered by their position in the path (though in this case they happen to be). If you do need them in order, you can do this (which is based on this answer about computing the position of elements in an RDF list):

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)

This finds each ancestor ?class of :Win7 as well as each ?mid intermediate ancestor. For ancestor ?class, the distance is computed as the number of intermediate relations in between (count(?mid)). It orders the results based that distance, so :Win7 is the closest ancestor, :Windows after that, and so on.

You can even do some of the fancy formatting you want like this:

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" |
-----------------------------------

It might be possible to do some fancier string processing and to get the multiline string. You might look at the latter part of this answer where there is some fancy formatting for a nicely aligned matrix for ideas.

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