Re: [xsl] generate-id accross different xsl files

[Francis Norton <francis@xxxxxxxxxxx>]

Jeni Tennison wrote:
> 
> 
> If there's no way of generating something unique from the information
> available around the node, then you can use the fact that it has a
> unique position within the XML document to give you the ID.  Two ones
> that I might use are the following:
> 
>  - element name + count of previous elements with the same name
> 
>   <xsl:variable name="name" select="name()" />
>   <xsl:value-of select="concat($name, '-',
>                                count(preceding::*[name() = $name])" />
> 

But 

"the preceding axis contains all nodes in the same document as the
context node that are before the context node in
document order, <em>excluding any ancestors</em> and excluding attribute
nodes and namespace nodes" (my <em/>)

so ...

---
C:\xml>type t.xml
<?xml version="1.0" encoding="UTF-8"?>
<a>
  <a>
    <a/>
  </a>
</a>

C:\xml>type t.xslt
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
  <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
  <xsl:template match="*">
    <xsl:copy>
      <xsl:attribute name="href">
        <xsl:variable name="name" select="name()"/>
        <xsl:value-of select="concat($name, '-',
count(preceding::*[name()= $name]))"/>
      </xsl:attribute>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

C:\xml>saxon t.xml t.xslt
<?xml version="1.0" encoding="UTF-8"?>
<a href="a-0">
   <a href="a-0">
      <a href="a-0"/>
   </a>
</a>
---

But 

"NOTE: The ancestor, descendant, following, preceding and self axes
partition a document (ignoring attribute and namespace nodes): they do
not overlap and together they contain all the nodes in the document."

So

	select="concat($name, '-', count(preceding::*[name()= $name]) +
count(ancestor::*[name()= $name]))"

should do it.

Sometimes even noble Homer nods ...  ;)

Francis.

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