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Hello. I'm working on a xml transformation to RDF with XSLT help. I'm totally very new to XSLT. And It's hard and I'm stuck now. I need your help. Here are xml element representation with three ways. <gemq:teachingMethods xsi:type='dcterms:URI'>http://purl.org/gem/instance/GEM-TM/#PeerTutoring</gemq:teachingMethods> <gemq:teachingMethods xsi:type='gemq:GEM-TM'>Thematic approach</gemq:teachingMethods> <gemq:teachingMethods>MultimediaInstruction</gemq:teachingMethods> I need to transform this to rdf like this With xsi:type="dcterms:URI" <rdf:Description> <gemq:teachingMethods rdf:resource="http://purl.org/gem/instance/GEM-TM/#PeerTutoring"/> </rdf:Description> With xsi:type="gemq:GEM-TM"
<rdf:Description>
<gemq:teachingMethods>
<gemq:GEM-TM>
<rdf:value>Thematic approach</rdf:value>
<rdfs:label>Thematic approach</rdfs:label>
</gemq:GEM-TM>
</gemq:teachingMethods>
</rdf:Description>And with just free-text <rdf:Description> <gemq:teachingMethods>MultimediaInstruction</dc:identifier> </rdf:Description> I did xslt with <xsl:template match="gemq:teachingMethods"> <xsl:choose> <xsl:when test="contains(@xsi:type, 'dcterms:URI')"> <xsl:attribute name="rdf:resource"> <xsl:value-of select="normalize-space(.)"/> </xsl:attribute> </xsl:when> <xsl:when test="not[@xsi:type]"> <xsl:attribute name="rdf:resource"> <xsl:value-of select="normalize-space(.)"/> </xsl:attribute> </xsl:when> <xsl:otherwise> <gemq:teachingMethods> <xsl:value-of select="normalize-space(.)"/> </gemq:teachingMethods> </xsl:otherwise> </xsl:choose> </xsl:template> Could you give me some tips or advice? I appreciate it. Thanks a lot, Best regards, Oknam
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