This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
Reverse order
- To: <XSL-List at lists dot mulberrytech dot com>
- Subject: [xsl] Reverse order
- From: Janning Vygen <vygen at planwerk6 dot de>
- Date: Tue, 1 May 2001 15:11:33 +0200
- Organization: Planwerk 6 /websolutions
- Reply-To: xsl-list at lists dot mulberrytech dot com
i am trying to list the parts of a book in reverse order.
<book>
<part><title>foo</title></part>
<part><title>bar</title></part>
<part><title>huhu</title></part>
</book>
<xsl:for-each select="/book/part[position()=last()]/preceding-sibling::part">
<xsl:value-of select="title"/><br/>
<xsl:for-each>
huhu<br>
bar<br>
foo<br>
I thought the XPath expression in the for-each statement should return a
reverse oder of all parts because it goes to the last part and looks back to
all preceding-siblings which are parts.
But it doesnt work, so i actually know that XPath doesnt 'look' the way i do
:-)
Is there a way to get a node set in a reverse order?
thanks in advance
janning
--
Planwerk 6 /websolutions
Herzogstraße 86
40215 Düsseldorf
fon 0211-6015919
fax 0211-6015917
http://www.planwerk6.de
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list