[xsl] ancestor axis ordering (again...)

Subject: [xsl] ancestor axis ordering (again...)
From: Joern Nettingsmeier <nettings@xxxxxxxxxxxxxxxxxxxxxx>
Date: Mon, 27 Feb 2006 14:43:45 +0100
hi *!

this template

<xsl:template match="d">
  ancestor axis: <xsl:for-each select="ancestor::*">
    <xsl:value-of select="name()"/>

last element: <xsl:value-of select="name(ancestor::*[last()])"/>


will convert this source



ancestor axis: abc

last element: a


the spec says that the ancestor axis is a reverse axis. but then the for-each statement would be wrong. from my understanding, it should produce "cba".

i found this explanation http://www.biglist.com/lists/xsl-list/archives/200405/msg00055.html
in the archive, but i don't understand the concept of "steps".
can anyone explain? sorry if this is a faq.



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