Subject: RE: [xsl] Returning a Tree From: "Sullivan, Dan" <dsullivan@xxxxxxxxxxx> Date: Tue, 4 Sep 2001 12:45:20 -0700 |
A variable without a select attribute contains a result tree fragment, which in effect is a string. There is no way to assign the result of a call-template with the select attribute... so in effect call-template always returns a result tree fragment, never a nodeset. So there is no standard way to do what you are trying to do. Many XSLT processors, however, have extension functions that will convert a result tree fragment into a nodeset. Dan -----Original Message----- From: Darren Hayduk [mailto:dhayduk@xxxxxxxxxxxxxxx] Sent: Tuesday, September 04, 2001 3:28 PM To: 'XSL-List@xxxxxxxxxxxxxxxxxxxxxx' Subject: [xsl] Returning a Tree I'm trying to construct a template (call-template) that will return a node set, and I can't figure out how to do it. I've had success calling a template with a node-set (a, below), assigning a node-set to a variable which can be used in the same context (b, below), but not returning a node-set (c, below). A: <xsl:call-template name="a"> <xsl:with-param select="//nodes" name="in"/> </xsl:call-template> B: <xsl:variable name="v" select="//nodes"/> C: <xsl:template name="c"> <xsl:value-of>... -or- <xsl:copy>...</xsl:template> and then <xsl:variable name="vc"> <xsl:call-template name="c"> </xsl:variable> <xsl:for-each select="$vc"> ... Is this possible?? Thanks, Darren _______________________________________________ Darren Hayduk | Network Management Nauticus Networks, Inc. 200 Crossing Blvd, Framingham, MA 01702 508-270-0500 x299 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] Returning a Tree, Michael Kay | Thread | [xsl] Unique processing, Darren Hayduk |
[xsl] Returning a Tree, Darren Hayduk | Date | Re: [xsl] default meta tag Revisted, David Carlisle |
Month |