Subject: Re: [xsl] Sorting a NodeSet Contained Within a Variable From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 11 Jul 2001 18:22:01 +0100 |
> In other words, I want to put those results into a nodeset and sort them. As must be stated in the FAQ you can not do this in XSLT 1.0 without use of a node-set() extension function. Any use of xsl:variable with content (as opposed to select attribute) does not construct a node st but a result tree fragment. You can not query into the structure of a result tree fragment at all, all you can do is convert it to a string, or copy it to the result. Most systems have an extension function ext:node-set() that allows you to convert the result tree fragment into a node set (containing a single node, a root node correspnding to the root pf the tree constructed by xsl:variable) David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Scanning Service. For further information visit http://www.star.net.uk/stats.asp or alternatively call Star Internet for details on the Virus Scanning Service. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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