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RE: [xsl] mixed content nodes question

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Hi Jeb,

> From: Jeb Boniakowski [mailto:jeb@xxxxxxxxxxx]
>
> <...snip...>
> I'd imagine I'd have some part of my xsl sheet include a long 'when'
> block and switch on the lookup attribute's values so i'd have:
> <xsl:when test="3">Foo</xsl:when>
> <xsl:when test="4">Bar</xsl:when>
>
> But I can't figure out how to do the actual transforming of only those
> <dictionary> tags in place, leaving the rest of the file unchanged.
> I've tried apply-templates in the a template matching "/", then having
> a template matching "*" do a test to see if the current node is
> <dictionary>, if so, transform and write out, otherwise, write out as
> it is, but I can't get it to work right.
>
> If it's vastly easier to process if instead of having:
> <dictionary lookup="3"/> -> Foo
> I have:
> <dictionary>3</dictionary> -> Foo
>
> I might be able to make that change to the xml format, but I'm still
> confused about how to deal with mixed nodes.  I feel like people must
> do this all the time, if they have bits of HTML embedded in larger XML
> trees they must have markup tags like <b>, <a> that make the nodes
> mixed content

Indeed... all the time.

First of all, you need a template like this:

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

Later, read http://www.dpawson.co.uk/xsl/sect2/defaultrule.html to learn
what you are fixing with the above template :-).

I put that template in its own stylesheet and then do something like

	<xsl:import href="xlst/common/identity.xslt" />

from all my other stylesheets that need this.  But for now, you can just
write the template inline.

Then, add a template like this:

	<xsl:template match="dictionary">
	   <xsl:if test="@lookup = '3'">Foo</xsl:if>
	   <xsl:if test="@lookup = '4'">Bar</xsl:if>
	<xsl:/template>

The above template uses the "@" shorthand notation for the attribute:: axis,
which you can read up on later.

HTH,
Mark

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