Re: [xsl] name of a template

[Jeni Tennison <mail@xxxxxxxxxxxxxxxx>]

Hi Carmelo,

>   Is it possible to simulate the equivalent of an enumerated list in XSLT
> (or any other list), then traverse the list and find the attributes that
> way?,
> Am I climbing the wrong tree by looking at that approach?

Ahh, no, you're onto something there :)

If you have your attribute sets defined as normal:

<xsl:attribute-set name="property1">
  <xsl:attribute name="foo">...</xsl:attribute>
  ...
</xsl:attribute-set>

<xsl:attribute-set name="property2">
  ...
</xsl:attribute-set>

Then you can access them through the document() function.  To get at
the 'property1' attribute set, for example, you can use:

  document('')/*/xsl:attribute-set[@name = 'property1']

Now because you've got the name as a string, then if you have stored
the position of the current node in a variable:

  <xsl:variable name="position" select="position()" />

then you can access the relevant attribute set using:

  document('')/*/xsl:attribute-set
    [@name = concat('property', $position)]

You can then iterate over the attributes in this attribute set, adding
the attributes:

  <xsl:for-each select="document('')/*/xsl:attribute-set
                          [@name = concat('property',
                                          $position)]/xsl:attribute">
    <xsl:attribute name="@name">
       <xsl:value-of select=".">
    </xsl:attribute>
  </xsl:for-each>

The big problem with this solution is that it won't work if the
attribute in the attribute set is defined with anything but a simple
value: if you have an xsl:choose or whatever to decide what the value
is, then the above won't work.

Nice idea :)

Cheers,

Jeni

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



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