如何用xslt查找一样的rdf:ID然后group成一个block，比如下面这个
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf
rdf:resource="#control_structure_performance_analysis_and_design_aids"/>
</rdf:Description>
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf rdf:resource="#performance_analysis_and_design_aids"/>
</rdf:Description>
正确的应该是
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf
rdf:resource="#control_structure_performance_analysis_and_design_aids"/>
<rdfs:subClassOf rdf:resource="#performance_analysis_and_design_aids" />
</rdf:Description>
而且原来的file里如果有重复的facts比如下面这个
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf
rdf:resource="#control_structure_performance_analysis_and_design_aids"/>
</rdf:Description>
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf
rdf:resource="#control_structure_performance_analysis_and_design_aids"/>
</rdf:Description>
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf rdf:resource="#performance_analysis_and_design_aids"/>
</rdf:Description>
正确的仍然是
<rdf:Description rdf:ID="simulation">
<rdf:type rdf:resource="http://www.w3.org/2000/01/rdf-schema#Class" />
<rdfs:subClassOf
rdf:resource="#control_structure_performance_analysis_and_design_aids"/>
<rdfs:subClassOf rdf:resource="#performance_analysis_and_design_aids" />
</rdf:Description>
应该怎么处理呢。。。什么思路阿谢谢
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2006/3/24 11:45:00
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[讨论]如何用xslt中找相同元素并进行以下操作
