Re: [xsl] determining position of a document included using document function

["Gennady Loskutov" <g_loskutov@xxxxxxxxxxxxx>]

> is there anyway to find out the position of file in the root document,
> without trying to do <xsl:apply-templates>
> <xsl:with-param name="filepos" select="position()"/>
> </xsl:apply-templates>

Bryan,

I use generate-id() function to identify the included file. Basically, the
idea is to match generated id for the root element of the current node with
ones, generated separately and stored in a variable. According to XSLT spec,
generate-id(document("foo.xml"))=generate-id(document("foo.xml")) is always
true, so I assume that
generate-id(document("foo.xml"))=generate-id(document(/files/file/@href))
must be also true if the "foo.xml" file is among those specified in
/files/file/@href.

To use my solution, you must use an extension function to transform a result
tree fragment stored in a variable into a node set (I use Microsoft XML
parser, which has node-set() function).

For example, using the following XML file

<?xml version="1.0" ?>
<files>
   <file href="data1.xml" />
   <file href="data2.xml" />
   <file href="data3.xml" />
</files>

I declare a variable, which is a union of the documents
<xsl:variable name="merged" select="document(/files/file/@href)" />

Then I store generated id for each document root in another variable:

<xsl:variable name="tops">
 <xsl:for-each select="/files/file">
  <file>
   <xsl:copy-of select="@href" />
   <xsl:attribute name="id"><xsl:value-of
select="generate-id(document(@href))" /></xsl:attribute>
  </file>
 </xsl:for-each>
</xsl:variable>

When I need to know the file URL, I do the following:
<xsl:template match="myElement">
...
 <xsl:variable name="currentTop" select="generate-id(/)" />
 <url>
  <xsl:value-of select="msxsl:node-set($tops)/file[@id = $currentTop]/@href"
/>
 </url>

Regards,
Gennady


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