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RE: [xsl] Find a specific element or fall back to element 1

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Try

select="(image[1] | image[@default='true'])[last()]"

Michael Kay
http://www.saxonica.com/ 

> -----Original Message-----
> From: John Horner [mailto:Horner.John@xxxxxxxxxx] 
> Sent: 15 January 2007 22:50
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] Find a specific element or fall back to element 1
> 
> I have a gallery.xml file containing a number of images:
> 
> <gallery>
>     <image>
>       <filename>1.jpg</filename>
>     </image>
>     <image>
>       <filename>2.jpg</filename>
>     </image>
>     <!-- etc -->
>     <image default="true">
>       <!-- optional attribute applies to this element -->
>       <filename>10.jpg</filename>
>     </image>
> </gallery>
> 
> and my xslt has to do the following thing -- look for an 
> image marked with a "default" attribute, and if it's not 
> found, use the first image instead. So I'm looking for the 
> position number of the marked element, and if I can't find 
> it, using 1:
> 
> <xsl:variable name="gallery-xml-file"
> select="document($gallery-xml-path)"/>
> <xsl:variable name="test-position">
>   <xsl:for-each select="$gallery-xml-file//image">
>     <xsl:if test="@default='true'">
>       <xsl:value-of select="position()"/>
>     </xsl:if>
>   </xsl:for-each>
> </xsl:variable>
> <xsl:variable name="display-position">
>   <xsl:choose>
>     <xsl:when test="$test-position = ''">1</xsl:when>
>     <xsl:otherwise>
>       <xsl:value-of select="$test-position"/>
>     </xsl:otherwise>
>   </xsl:choose>
> </xsl:variable>
> 
> This seems a bit long and wasteful to me. Is there a better way?
> 
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