[XSL-LIST Mailing List Archive Home] [By Thread] [By Date]

Re: [xsl] Copying Namespace Nodes

---

---

At 2011-11-27 22:10 +0100, Heiko Niemann wrote:

I have an element I want to copy and in the source it looks like this:

<a:elem
  xmlns:a="example.com/ns/a"
  xsi:type="nsc:foo"
  xmlns:nsc="example.com/ns/sensitive"
  xmlns:b="example.com/ns/b"
  xmlns:c="example.com/ns/c"
  ...
  xmlns:z="example.com/ns/z">info</a:elem>

What I want to achieve at the same time:

1) Keep the namespace declaration for the namespace sensitive content of
the xsi:type attribute (with prefix nsc).

2) Get rid of all namespace declarations that are not needed here (prefix
b, c, ..., z) and that clutter up my result document.

If I use copy-of with copy-namespaces set to 'no' the declaration for
namespace sensitive content will be lost. If I say 'yes' everything stays
the same. So is there a short way I have not seen yet or is the solution
some verbose template? :)

I don't think the template needed is very verbose at all since you are limiting yourself to finding the one prefix that is found in the xsi:type= attribute:

T:\ftemp>type heiko.xml
<?xml version="1.0" encoding="UTF-8"?>
<a:elem
xmlns:a="example.com/ns/a"
xsi:type="nsc:foo"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:nsc="example.com/ns/sensitive"
xmlns:b="example.com/ns/b"
xmlns:c="example.com/ns/c"
xmlns:z="example.com/ns/z">info</a:elem>
T:\ftemp>xslt2 heiko.xml heiko.xsl
<?xml version="1.0" encoding="UTF-8"?><a:elem xmlns:a="example.com/ns/a" xmlns:nsc="example.com/ns/sensitive" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="nsc:foo"/>
T:\ftemp>type heiko.xsl
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
version="2.0">

<xsl:template match="*">
  <xsl:copy copy-namespaces="no">
    <xsl:for-each select="@xsi:type">
      <xsl:copy-of
          select="../namespace::*[name(.)=substring-before(current(),':')]"/>
      <xsl:copy/>
    </xsl:for-each>
  </xsl:copy>
</xsl:template>

</xsl:stylesheet>
T:\ftemp>

I hope this helps. If there is no xsi:type attribute, then no copy of the namespace node is done.

. . . . . . . . Ken

Thanks for help,
Heiko

--
Contact us for world-wide XML consulting and instructor-led training
Free 5-hour video lecture: XSLT/XPath 1.0 & 2.0 http://ude.my/t37DVX
Crane Softwrights Ltd.            http://www.CraneSoftwrights.com/s/
G. Ken Holman                   mailto:gkholman@xxxxxxxxxxxxxxxxxxxx
Google+ profile: https://plus.google.com/116832879756988317389/about
Legal business disclaimers:    http://www.CraneSoftwrights.com/legal

---