How to do an XSL:for-each in reverse order

Question

I am looking to reverse in XSL/FO a for-each loop.

for instance the xml

<data>   <record id="1"/>   <record id="2"/>   <record id="3"/>   <record id="4"/>   <record id="5"/>   <record id="6"/> </data> 

with the xsl

<xsl:for-each select="descendant-or-self::*/record">    <xsl:value-of select="@id"/> </xsl:for-each> 

I am looking for the output 654321 and not 123456

how is this possible?

Answer 1

Use xsl:sort not for ordering by @id but for ordering by position():

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template match="/data">     <xsl:for-each select="descendant-or-self::*/record">         <xsl:sort select="position()" data-type="number" order="descending"/>         <xsl:value-of select="@id"/>     </xsl:for-each> </xsl:template> </xsl:stylesheet> 

Answer 2

Yes, Alexander is right - forgot the data-type though:

<xsl:for-each select="descendant-or-self::*/record">    <xsl:sort select="@id" order="descending" data-type="number" />    <xsl:value-of select="@id"/> </xsl:for-each> 

(without that, you'll run into sorting problems with numbers over 9)