XSLT: Select nodes in reverse order

[thrill5]

I have an xml document such as:
<device>
<element>first</element>
<element>second</element>
</device>

I am using this as the source for an xslt transform that goes like

<xsl:for-each select="/device/element>
This is the <xsl:value-of select="."/> element.
</xsl:for-each>

This would result in the following:

This is the first element
This is the second element

Is it possible to select the nodes in reverse order so that I would end up
with:

This is the second element
This is the first element

I can add a unique attribute to <element> if that would help such as:
<device>
<element id="1">first</element>
<element id="2">second</element>
</device>

Is there an XPath statement I can use in the <xsl:for-each> to select the
elements in reverse order using the attribute as an order tag?

Scott

[Jeff Higgins]

thrill5 wrote:

I have an xml document such as:
<device>
<element>first</element>
<element>second</element>
</device>

I am using this as the source for an xslt transform that goes like

<xsl:for-each select="/device/element>
This is the <xsl:value-of select="."/> element.
</xsl:for-each>

This would result in the following:

This is the first element
This is the second element

Is it possible to select the nodes in reverse order so that I would end up
with:

This is the second element
This is the first element

I can add a unique attribute to <element> if that would help such as:
<device>
<element id="1">first</element>
<element id="2">second</element>
</device>

Is there an XPath statement I can use in the <xsl:for-each> to select the
elements in reverse order using the attribute as an order tag?

Scott

Scott,
I'm real new at this so take it with a grain of salt.

<xsl:template match="//device">
<xsl:for-each select="element">
<xsl:sort select="@id" data-type="number" order="descending"/>
<xsl:text>This is the </xsl:text>
<xsl:value-of select="."/>
<xsl:text> element.</xsl:text>
</xsl:for-each>
</xsl:template>

[Joe Kesselman]

Jeff Higgins wrote:

I'm real new at this so take it with a grain of salt.

You were on the right track...

The nodes selected by xsl:for-each are normally presented in document
order. To get reverse document order, you want to reverse that
positioning -- which can be done by re-sorting in descending order by
original position.

<xsl:template match="/">
<xsl:for-each select="/device/element">
<xsl:sort select="position()" data-type="number" order="descending"/>
This is the <xsl:value-of select="."/> element.
</xsl:for-each>
</xsl:template>

--
() ASCII Ribbon Campaign | Joe Kesselman
/\ Stamp out HTML e-mail! | System architexture and kinetic poetry

[Jeff Higgins]

Joe Kesselman wrote:

... To get reverse document order, you want to reverse that positioning --
which can be done by re-sorting in descending order by original position.

<xsl:sort select="position()" data-type="number" order="descending"/>

How neat!
Thanks, Joe.