Easy in XPath 2.0:
for $i in 1 to count(BB/BB1)
return (BB/BB1[$i], AA/AA1[$i])
Just slightly more verbose in XSLT 1.0:
<xsl:variable name="root" select="."/>
<xsl:for-each select="BB/BB1">
<xsl:variable name="i" select="position()"/>
<xsl:copy-of select="."/>
<xsl:copy-of select="$root/AA/AA1[$i]"/>
</xsl:for-each>
Michael Kay
-----Original Message-----
From: Anthony Ettinger [mailto:apwebdesign(_at_)yahoo(_dot_)com]
Sent: 07 September 2004 18:47
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] xsl needed: sibilings with different parents
Hi,
I've the following document:
<root>
<AA>
<AA1/>
<AA1/>
<AA1/>
......
</AA>
<BB>
<BB1/>
<BB1/>
<BB1/>
......
</BB>
</root>
I need to loop on BB/BB1 and within the loop, I need
to use the index or position of BB/BB1 and access
AA/AA1 to get the corresponding AA/AA1's value each
time.
Any thoughts?
Thank you
=====
Anthony Ettinger
Phone: (408) 656-2473
apwebdesign(_at_)yahoo(_dot_)com
http://www.apwebdesign.com
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