Re: [xsl] XSLT template from XSLT + XML

[Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>]

Hi Michael,

> Just out of curiosity, is it possible to create a stylesheet that
> will go in and recreate all elements and attributes? I can see it
> being possible for the elements, but the attributes would be tough,
> unless you can do a for-each select="attributes" type of thing.

Sure. The easiest "identity" stylesheet is:

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

<xsl:template match="/">
  <xsl:copy-of select="." />
</xsl:template>

</xsl:stylesheet>

But that just copies everything exactly as it was. If you want to make
a few adjustments, you need to use an "identity template", which you
can override for specific elements and attributes by defining
templates for those elements and attributes:

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

<xsl:template match="node() | @*">
  <xsl:copy>
    <xsl:apply-templates select="@* | node()" />
  </xsl:copy>
</xsl:template>

... your overriding templates here ...

</xsl:stylesheet>

You can also copy elements and attributes using xsl:element and
xsl:attribute with attribute value templates in the name and namespace
attributes:

  <xsl:element name="{name()}" namespace="{namespace-uri()}">
    ...
  </xsl:element>

  <xsl:attribute name="{name()}" namespace="{namespace-uri()}" />

though generally xsl:copy is easier.
  
Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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