Question

I have a simple HTML fragment similar to this:

<a href="123">link</a>

I need to transform it to

<abc:href var="123">link</abc:href>

I do it with XSLT, so I had to add the namespace in xsl:stylesheet

<xsl:stylesheet version="1.0"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:abc="http://abc.ru">

It works almost fine, unfortunately the XSLT transform keeps on adding a XMLNS to the output, like here:

<abc:href var="123" xmlns:abc="http://abc.ru">link</abc:href>

I don’t need the xmlns definition, can I remove it?

Answer 1

Although it really goes against the grain, and I advise strongly against it, if you need to produce this malformed XML, then you can use an instruction like…

<xsl:value-of disable-output-escaping="yes" select="
  concat('&lt;abc:href var="',$href,'"&gt;',$link,'&lt;/abc:href&gt;')
 "/>

… where $href and $link are place-markers for the appropriate expression.

---

Update

In response to the OP’s comment, one could use a template like this…

<xsl:template match="a">
  <xsl:value-of disable-output-escaping="yes" select="
    concat('&lt;abc:href var="',@href,'"&gt;',.,'&lt;/abc:href&gt;')
   "/>
</xsl:template>

This ugly solution should be used only as a last resort. A much better solution would be to use XSLT to produce your WHOLE document, not just an invalid fragment of it. This way you document would be well formed and you could bring to bear the full power and simplicity of XSLT.