Re: [xsl] convert xml with schema to html with xslt

[Mukul Gandhi <gandhi.mukul@xxxxxxxxx>]

It looks like, that your XML document is in a namespace (defined by,
xmlns="urn:Verlag").

But your stylesheet doesn't consider namespaces.

for example, this template:
<xsl:template name="paragraph" match="para">

will not match the "para" node from XML, as "para" node in your XML is
in a namespace (which is, "urn:Verlag").

Importing the schema in the stylesheet won't solve this problem,
because schema import in XSLT 2.0 is for a different need (basically
to provide the schema types and declarations to the stylesheet).

I think, something like below may solve this problem:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:ns1="urn:Verlag" version="2.0">

  <xsl:template name="paragraph" match="ns1:para">
  ...
</xsl:stylesheet>

On Mon, Dec 21, 2009 at 3:51 PM, Lajos Joo <lajosjoo@xxxxxxxxx> wrote:
> Hello!
>
> I have a problem. I have a small xml file which has a schema declaration in
the root element. I cannot convert it to html with an xslt file. If i remove
the schema declaration the conversion is fine.
> Please suggest me how to correct this.
> I have included a shcema import but it didnt help:
> <xsl:import-schema namespace="urn:Verlag" schema-location="urn:Verlag
http://192.168.190.181:8879/xml/STRUKTUR.xsd"; />
>
> The xml file looks like this:
> <LOSCHNIGG xmlns="urn:Verlag"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance";
xsi:schemaLocation="urn:Verlag http://192.168.190.181:8879/xml/STRUKTUR.xsd";>
> B  B  B  B <metadaten>
> B  B  B  B  B  B  B  B <gilt-ab>1000-01-01</gilt-ab>
> B  B  B  B  B  B  B  B <gilt-bis>9999-12-31</gilt-bis>
> B  B  B  B  B  B  B  B <Fassung_Publikation>
> B  B  B  B  B  B  B  B  B  B  B 
B <Publikationsstand>2009-12-19</Publikationsstand>
> B  B  B  B  B  B  B  B  B  B  B  B <Release_Version>1</Release_Version>
> B  B  B  B  B  B  B  B </Fassung_Publikation>
> B  B  B  B </metadaten>
> B  B  B  B <para>
> text...
> B  B  B  B </para>
> </LOSCHNIGG>
>
> The xml works fine and the xsd is in the right place.
>
> The xsl looks like this:
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns="http://www.w3.org/1999/xhtml"; version="2.0">
>
> <!-- B  B <xsl:import-schema namespace="urn:Verlag"
schema-location="urn:Verlag http://192.168.190.181:8879/xml/STRUKTUR.xsd";
/>-->
>
> B  B  B  B <xsl:output encoding="UTF-8" indent="yes" method="html"
version="1.0" />
>
> B  B  B  B <xsl:template match="@*|node()">
> B  B  B  B  B  <xsl:copy>
> B  B  B  B  B  B  B <xsl:apply-templates select="@*|node()"/>
> B  B  B  B  B  </xsl:copy>
> B  B  B  B </xsl:template>
>
> B  B  B  B  B  B  B  B <xsl:template name="paragraph" match="para">
> B  B  B  B  B  B  B  B  B  B  B  B <xsl:element name="p">
> B  B  B  B  B  B  B  B  B  B  B  B  B  B  B  B <xsl:apply-templates/>
> B  B  B  B  B  B  B  B  B  B  B  B </xsl:element>
> B  B  B  B  B  B  B  B </xsl:template>
>
> B  B  B  B <xsl:template match="/">
> B  B  B  B  B  B  B  B <xsl:element name="html">
> B  B  B  B  B  B  B  B  B  B  B  B <xsl:element name="head">
> B  B  B  B  B  B  B  B  B  B  B  B  B  B  B  B <xsl:element
name="title">Preview</xsl:element>
> B  B  B  B  B  B  B  B  B  B  B  B </xsl:element>
> B  B  B  B  B  B  B  B  B  B  B  B <xsl:apply-templates/>
> B  B  B  B  B  B  B  B </xsl:element>
> B  B  B  B </xsl:template>
> </xsl:stylesheet>
>
> Thank you!
> Lajos



--
Regards,
Mukul Gandhi