RE: [xsl] Xsl:copy-of writes xmlns:xsi always - Any way to avoid this?

Subject: RE: [xsl] Xsl:copy-of writes xmlns:xsi always - Any way to avoid this?
From: <binu.idicula@xxxxxxxxx>
Date: Mon, 5 Mar 2007 15:26:34 +0530
Thanks  Micheal for the quick reply,

I have to stick onto XSLT1.0.

I used the following .. But the problem was that it is not copying
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

            <xsl:for-each select="./*">
                <xsl:element name="{name()}"
                <xsl:copy-of select="@*"/>

Once I select a particular node, all the contents (including next level
- children) should be translated to another XML. Please advice.

Binu Kuttikkattu Idicula

-----Original Message-----
From: Michael Kay [mailto:mike@xxxxxxxxxxxx]
Sent: Monday, March 05, 2007 2:53 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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=" 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

<xsl:template match="*">
  <xsl:element name="{name()}" namespace="{namespace-uri()}">
  <xsl:copy-of select="@*"/>

Michael Kay

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