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Re: How to handle xsi prefix in XSLT

Subject: Re: [xsl] How to handle xsi prefix in XSLT
From: "Thomas B. Passin" <tpassin@xxxxxxxxxxxx>
Date: Fri, 30 Nov 2001 10:17:27 -0500
xslt example template for xsi
Using your original xml sample, and this stylesheet, this version of the
transformation works fine using either msxml3 or Saxon.  You are going to
have to tell us what system and processors you use, and how you are trying
to produce the transformation, and how you are trying to use it.  For
example, if you expect the processor to find the stylesheet from the
processing instruction in the xml file, that may not be working for a number
of reasons.

BTW, your method for producing a "Root"  element works fine, but it is a
little easier to get the same result like this:

<xsl:template match='/'>
   <Root>
       <xsl:apply-templates/>
   </Root>
</xsl:template>

Tom P

[Janusz Dalecki]

Please forgive my ignorance, I am new to XSLT, I have fixed my XSLT to the
one
below which inserts a root element (“Root”) and I am still getting the same
error – am I doing something stupid?

<?xml version = "1.0" encoding = "UTF-8"?>
<xsl:stylesheet xmlns:xsl = "http://www.w3.org/1999/XSL/Transform" xmlns:xsi
=
"http://www.w3.org/2000/10/XMLSchema-instance" version = "1.0">

<xsl:template match = "/">
 <xsl:element name = "Root">
  <xsl:apply-templates/>
 </xsl:element>
</xsl:template>

<xsl:template match = "Command">
 <xsl:element name = "Transaction">
  <xsl:apply-templates select = "@*"/>
 </xsl:element>
</xsl:template>


<xsl:template match = "Command[@xsi:type]">
 <xsl:element name = "hhhhhhh">
 </xsl:element>
</xsl:template>

</xsl:stylesheet>




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