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[xsl] Reverse order


Subject: [xsl] Reverse order
From: Janning Vygen <vygen@xxxxxxxxxxxx>
Date: Tue, 1 May 2001 15:11:33 +0200

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

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