Subject: RE: [xsl] constructing the Node Sets From: "Roger Glover" <glover_roger@xxxxxxxxx> Date: Tue, 28 Jan 2003 00:39:11 -0600 |
Idea copied from Michael Kay's "XSLT Programmer's Reference", 2nd ed., Wrox Press, 2001, p. 425 ... <xsl:variable name="C" select="$A[ count( . | $B ) = count( $B ) ]"/> The select expression selects the node subset C of node set A, composed of any member node of A which, when added to node set B, produces a node set the same size as B (in other words, the node was already in node set B). The usage of A and B in the select expression can be reversed with no side effects except possibly: o a change in the list ordering of nodes in node set C o a change in the speed of execution of the expression -- Roger Glover Siva Jasthi wrote > I have constructed two variables (A and B) each of which contains a Node > Set (through select="XPath Expression"). > > A = [Node1, Node2, Node3, Node4, Node5] > B = [Node3, Node4, Node5, Node6, Node7, Node8] > > I now would like to construct another variable C with the Nodes that exist > both in A and B. (So as to produce a third table with the Nodes that are > common to both A and B). > > C= [Node3, Node4, Node5] > > Is there any way to construct C from A and B?. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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