Would it be like
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:apply-templates select="@*|node()" />
</xsl:template>
<xsl:template match="connectivity[not(*)]">
<xsl:copy>
<xsl:apply-templates select="@*|node()" />
</xsl:copy>
</xsl:template>
regards,
Ranjan
--- Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:
Your template is copying a node only if it has one
or more children named
Entry.
Use a standard identity template and supplement it
with the second rule:
<xsl:template match="Entry[not(*)]"/>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Ranjan K. Baisak
[mailto:ranjanbaisak(_at_)yahoo(_dot_)com]
Sent: 15 June 2005 07:39
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Copy all nodes
I forgot to mention, if any <Entry> element with
zero
child, then I dont wan to copy.
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:if test="count(child::Entry)>0">
<xsl:apply-templates select="@*|node()" />
</xsl:if>
</xsl:copy>
</xsl:template>
The above template does not work
-Ranjan
--- "Ranjan K. Baisak" <ranjanbaisak(_at_)yahoo(_dot_)com>
wrote:
Hello,
I need a xpath expression to copy all nodes
from
source.
e.g. input: <Entry>
<Session>2004/5</Session>
<Department>Accounting and
Finance</Department>
</Entry>
output:<Entry>
<Session>2004/5</Session>
<Department>Accounting and
Finance</Department>
</Entry>
Can anybody tell me the xpath expression for the
same.
regards,
Ranjan
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