I used the following .. But the problem was that it is not
copying nested.
Yes, my code was right and yours is wrong, so please use mine.
Michael Kay
http://www.saxonica.com/
It just copies the direct chile element nodes. If the child
element has another child, do I have to write one more
for-each OR is there a simple way.
<xsl:for-each select="./*">
<xsl:element name="{name()}"
namespace="{namespace-uri()}">
<xsl:copy-of select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:for-each>
Once I select a particular node, all the contents (including
next level
- children) should be translated to another XML. Please advice.
regards
Binu Kuttikkattu Idicula
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: Monday, March 05, 2007 2:53 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Xsl:copy-of writes xmlns:xsi always - Any
way to avoid this?
Hi,
I use <xsl:copy-of select="./*"/> to select child
elements of a node
which passes <xsl:when>. However copy-of is inserting
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance with
each child
element it selects.
In XSLT 2.0 you can copy an element without copying its
namespaces using the copy-namespaces="no" attribute.
In XSLT 1.0 the answer is no. xsl:copy-of copies a tree
unchanged, which includes its namespace nodes (remember that
in the data model, an element has namespace nodes
corresponding to all in-scope namespaces, including those
declared on ancestor elements). You only remedy is not to use
xsl:copy-of, but to do a manual copy using a modified identity
template:
<xsl:template match="*">
<xsl:element name="{name()}" namespace="{namespace-uri()}">
<xsl:copy-of select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
Michael Kay
http://www.saxonica.com/
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