问题描述
我正在尝试将XML源映射到RDF,并且想知道如何在每次跟踪父级属性的同时进行递归遍历。在所示示例中,我希望提取所有Person
个节点,并将每个父节点的name
附加到当前节点。考虑所有父名称是必要的,因为节点名称可能不是唯一的。 (在示例中,两次访问grandChild1
)
请注意,我事先不知道嵌套可以进行多少个级别,因此为每个级别添加一个TriplesMap都不可行。
在经历了documentation,examples和test cases之后,我不确定这是否可能。
下面是一个简化的示例XML,我到目前为止创建的rml映射,使用RMLMapper生成的当前RDF输出以及我期望的RDF输出。
数据<Root>
<Person>
<name>parent1</name>
<Children>
<Person>
<name>child1</name>
<Person>
<name>grandchild1</name>
<Children>
<Person>
<name>greatgrandchild1</name>
</Person>
</Children>
</Person>
</Person>
<Person>
<name>child2</name>
<Person>
<name>grandchild1</name>
</Person>
</Person>
</Children>
</Person>
</Root>
制图
@prefix rml: <http://semweb.mmlab.be/ns/rml#> .
@prefix rr: <http://www.w3.org/ns/r2rml#> .
@prefix ql: <http://semweb.mmlab.be/ns/ql#> .
@prefix owl: <http://www.w3.org/2002/07/owl#> .
@prefix rdf: <http://www.w3.org/1999/02/22-rdf-Syntax-ns#> .
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
@prefix testont: <http://www.example.com/ontology/> .
@prefix : <http://www.example.com/rules/> .
@base <http://www.example.com/instance/> .
:TriplesMapPerson a rr:TriplesMap;
rml:logicalSource [
rml:source "recursion_data.xml";
rml:referenceFormulation ql:XPath;
rml:iterator "//Root/Person"
].
:TriplesMapPerson rr:subjectMap [
rr:template "{name}"
].
:TriplesMapPerson rr:predicateObjectMap [
rr:predicate rdf:type;
rr:object testont:Person
].
:TriplesMapChild a rr:TriplesMap;
rml:logicalSource [
rml:source "recursion_data.xml";
rml:referenceFormulation ql:XPath;
rml:iterator "//Root/Person/Children/Person"
].
:TriplesMapChild rr:subjectMap [
rr:template "{name}_{../../name}"
].
:TriplesMapChild rr:predicateObjectMap [
rr:predicate rdf:type;
rr:object testont:Person
].
当前的RDF
<http://www.example.com/instance/parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/child1_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/child2_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
预期的RDF
<http://www.example.com/instance/parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/child1_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/grandchild1_child1_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/greatgrandchild1_grandchild1_child1_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/child2_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
<http://www.example.com/instance/grandchild1_child2_parent1> <http://www.w3.org/1999/02/22-rdf-Syntax-ns#type> <http://www.example.com/ontology/Person>.
感谢您为解决此问题所提供的帮助。预先感谢!
解决方法
您可以利用XPath的//
运算符的递归性质对所有人员进行迭代,而不管其深度如何。
然后,在主题模板中,您可以使用XPath的“祖先或自身”轴来获取具有名称的祖先节点的列表,将其反向以获得所需的顺序,并获取每个节点的name属性:
for $ancestor in reverse(ancestor-or-self::*[name]) return $ancestor/name
然后,您可以使用XPaths字符串连接函数来连接值以获得所需的结果。
映射将如下所示
:Person a rr:TriplesMap ;
rml:logicalSource [
rml:source [a carml:Stream ];
rml:referenceFormulation ql:XPath;
rml:iterator "//Person"
] ;
rr:subjectMap [
rr:template "{string-join(for $ancestor in reverse(ancestor-or-self::*[name]) return $ancestor/name,'_')}" ;
rr:class testont:Person ;
] ;
.
给出以下结果:
<http://example.com/base/parent1> a <http://www.example.com/ontology/Person> .
<http://example.com/base/child1_parent1> a <http://www.example.com/ontology/Person> .
<http://example.com/base/grandchild1_child1_parent1> a <http://www.example.com/ontology/Person> .
<http://example.com/base/greatgrandchild1_grandchild1_child1_parent1> a <http://www.example.com/ontology/Person> .
<http://example.com/base/child2_parent1> a <http://www.example.com/ontology/Person> .
<http://example.com/base/grandchild1_child2_parent1> a <http://www.example.com/ontology/Person> .
请注意,要使其正常工作,映射工具的XPath实现需要支持反向轴遍历。我通过支持该示例的https://github.com/carml/carml运行了该示例。
还要注意,您的xml输入示例缺少某些Children
节点的某些中间Person
节点。