[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Why do the namespace appears in transformation ?
Hi List. I am trying to generate an XSL template from an XML file that describes some filters. So, basically, I get <Results> <Row> <MTF_NUMERO_TABLEAU>2</MTF_NUMERO_TABLEAU> <TA_TITRE_A>Titre 1</TA_TITRE_A> </Row> <Row> <MTF_NUMERO_TABLEAU>3</MTF_NUMERO_TABLEAU> <TA_TITRE_A>Titre 2</TA_TITRE_A> </Row> </Results> I am parsing it with (not finished, of course) <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" exclude-result-prefixes="xsl" xmlns:xso="dummy" > <xsl:output method = "xml" encoding="UTF-8"/> <!-- When transforming, all xso namespace elements will become xsl --> <xsl:namespace-alias stylesheet-prefix="xso" result-prefix="xsl"/> <xsl:template match="/Results"> <xsl:for-each-group select="Row" group-by="MTF_NUMERO_TABLEAU"> <xsl:variable name="numtableau"> <xsl:value-of select="current-grouping-key()"/> </xsl:variable> <xso:template match="node()[(MTF_NUMERO_TABLEAU = '{$numtableau}')]"> <!--Do something--> </xso:template> <xsl:value-of select="current-grouping-key()"/>, </xsl:for-each-group> </xsl:template> <xsl:template match="Row"> </xsl:template> </xsl:stylesheet> The output (with Kernow) is <?xml version="1.0" encoding="UTF-8"?> <xsl:template xmlns:xsl="http://www.w3.org/1999/XSL/Transform" match="node()[(MTF_NUMERO_TABLEAU = '2')]"/>2, <xsl:template xmlns:xsl="http://www.w3.org/1999/XSL/Transform" match="node()[(MTF_NUMERO_TABLEAU = '3')]"/>3 Which is almost what I want.... But, how can I get rid of the xmlns declaration in each <xsl:template section ? (So that to get <?xml version="1.0" encoding="UTF-8"?> <xsl:template match="node()[(MTF_NUMERO_TABLEAU = '2')]"/>2, <xsl:template match="node()[(MTF_NUMERO_TABLEAU = '3')]"/>3 ) Thanks in advance Regards, Fabien
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