Re: [xsl] problem with xsl:sort

Subject: Re: [xsl] problem with xsl:sort
From: David Carlisle <davidc@xxxxxxxxx>
Date: Fri, 21 Feb 2003 11:02:40 GMT

  > <xsl:for-each select=".">
  > <xsl:sort select="@type" data-type="text" order="ascending"/>
  > <xsl:for-each select="."> processes a node-set that always contains
  > exactly one node. So there isn't very much point in sorting it.

  What's the problem? Have I done something wrong ?

Not necessarily wrong, just pointless. If you sort a list of length 1
then clearly the result will be the same as the original list so
there is no point in sorting.

"." matches just the current node so

<xsl:for-each select=".">

will execute over a current node list which is just the current node

<xsl:for-each select=".">
<xsl:sort select="@type" data-type="text" order="ascending"/>

is exactly the same thing, whataver sort options you specify.


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