Array treatment in XSLT

Question

Guys,
I have a XML like:
<root>
<people>
<name>John</name>
<name>Peter</name>
</people>
<ID>
<personalID>001</personalID>
<personalID>002</personalID>
</ID>
</root>
I need to order these people's names and IDs in person tags in target XML, like:
<people>
<person>
<name>John</name>
<personalID>001</personalID>
</person>
<person>
<name>Peter</name>
<personalID>002</personalID>
</person>
</people>
It's easy to do this in Message Mapping. However, I need to do this with XSLT (in order to keep subnodes' indentation, since Message Mappings will remove any indentation on the source message).
I've thought a while about it but I didn't find any way of doing it with XSLT. I mean, I need to make a for-each loop and to copy-of the equivalent nodes (name and person) for that loop index. I tried to use position() but with no success. How can I get the node that corresponds to some index from a subset?
Thanks in advance,
Henrique.

Answer 1

Well,
it will be the same since the expression "$index" does not denote a proper condition (it's always true).
I have implemented a xslt with one loop only but with the proper condition expression. It's like this:
<?xml version='1.0'?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/">
<xsl:variable name="idRoot" select="root/ID/personalID"/>
<people>
<xsl:for-each select="root/people/name">
<person>
<xsl:copy-of select="."/>
<xsl:variable name="index" select="position()"/>
<xsl:copy-of select="$idRoot[position() = $index]"/>
</person>
</xsl:for-each>
</people>
</xsl:template>
</xsl:stylesheet>
Regards,
Henrique.