Re: [xsl] Question about xsl-xslt

[Dennis Thomas <dthomas@xxxxxxxxx>]

Hi Saverio,
The problem is in the defination of
     <xsl:variable name="level" select="3"></xsl:variable>
When u define the select="3" its looking for a element node  3.
Rather if u want to store a value u need to declare it as --> select="'3'"

The declaration would now look as
 <xsl:variable name="level" select="'3'"></xsl:variable>

take care,
dennis

Saverio Niccolini wrote:

> Dear all,
>
> I'm new in xml world so pease forgive me if my question is a silly one
> I'd like to define and test a variable in my xslt stylesheet but it
> didn't work:
>
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
>  <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
>  <xsl:template match="/">
>  <xsl:variable name="level" select="3"></xsl:variable>
>                  <xsl:choose>
>                     <xsl:when test="$level=3">
>                         <xsl:value-of select="/filevideo/campi/campo1">
>                         </xsl:value-of>
>                     </xsl:when>
>                  </xsl:choose>
> </xsl:template>
>
> It doesn't work because xslt doesn't recognize xsl:choose and xsl:when
> which belengs to xsl namespace?
> How may I make it work? I have to define two namespace in my xml
> document or what?
>
> Please any comment is well accepted
> Thanks in advance
> Saverio
>
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