Re: How to handle xsi prefix in XSLT

I was using visual tool called "TIBCO Software Inc." without much success.
I have downloaded XML Spy and it works - so it was a tool problem.
Thanks for your help and the time you spent answering my questions.
Janusz


"Thomas B. Passin" wrote:

> 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>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list