Subject: Re: [xsl] Can sets have order? From: David Carlisle <davidc@xxxxxxxxx> Date: Thu, 25 Jan 2001 17:47:37 GMT |
Just noticed you said: > > It just constructs the set of ancestors. > In a particular order, right? Do you claim that it can return a node set in > any order besides reverse document order? YES!!! If you want to claim that it returns a node set that is ordered, it is ordered in document order. <xsl:variable name="x" select="attribute::*"/> OK so $x is a node set. If it is ordered, then presumably <xsl:copy-of select="$x[1]"/> is it's first element, no? well it gets the outer most element, the relevant order being document order not reverse document order. If $x was an odered set ordered by reverse document order, then <xsl:copy-of select="$x[1]"/> would be the same as <xsl:copy-of select="ancestor::*[1]"/> and select the parent. It is in explaining why <xsl:copy-of select="$x[1]"/> isn't the same as <xsl:copy-of select="ancestor::*[1]"/> that it is helpful to stress that the node set $x is just a set, that it doesn't "remember" that it was constructed in reverse order. In other words the ordering isn't a property of the set $x. David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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