Re: [xsl] elements to attributes

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Hi Harsha,

At 02:23 PM 2/28/2003, you wrote:
> and I need my result XML document to have all the attributes in each level
> converted to elements at same level
> Can there be a generic xsl which looks at all the nodes and does this?

Uh-huh...

There's a template described in the XSL Recommendation 
(http://www.w3.org/TR/xslt) as an "identity template", because it matches 
any node and just copies it, then picks its attributes and children to do 
the same:

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

You need to split this in two: one to handle all nodes but attributes, the 
other to handle attributes:

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

Except the one for attributes should have instructions to make an element 
instead:

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

But if you try this out, you really should study up on the processing model 
(templates and apply-templates, and the default child:: axis) to make sure 
you understand it.

(Once you understand the technique, you also will know how to extend it, 
for example to move things around or change names -- or add things.)

The identity template (find it in the list archives under identity template 
or identity transformation, it's a FAQ too) is also the solution to the 
other problem you have posted. You may have already gotten a solution, or 
ask again if you need to and someone will give you a hint.

Cheers,
Wendell


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