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RE: Returning a Tree
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] Returning a Tree
- From: "Sullivan, Dan" <dsullivan at develop dot com>
- Date: Tue, 4 Sep 2001 12:45:20 -0700
- Reply-To: xsl-list at lists dot mulberrytech dot com
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@nauticusnet.com]
Sent: Tuesday, September 04, 2001 3:28 PM
To: 'XSL-List@lists.mulberrytech.com'
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
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