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XSLT General questions and answers about XSLT. For issues strictly specific to the book XSLT 1.1 Programmers Reference, please post to that forum instead.
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Old June 19th, 2006, 09:28 AM
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Default node-set function

Hi,
Is something like this possible?

I have an outside function that returns xml, and I want to store that xml in an xslt variable and further process it. For example:

Code:
<xsl:variable name="getXml">

</xsl:variable>

<xsl:value-of select="exsl:node-set($getXml)/element"/>
Is this possible? If not, what other options do I have? Thanks a lot!

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Old June 19th, 2006, 09:54 AM
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That's very possible, most processors have a node-set function. Alternatively the document function will return a document root so you won't need an extension:
Code:
 <xsl:variable name="externalDoc" select="document('http://www.myDomain.com/myXmlSite/myXmlPage.aspx')"/>
In version 2.0 of XSLT you don't need an extension function, even if your variable is like:
Code:
<xsl:variable name="externalDoc"></xsl:variable>
as this is converted automatically to a root node.

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Old June 19th, 2006, 10:45 AM
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Many XSLT processors support this kind of functionality, but the details depend on the processor.

If the external function returns a node-set, there will be no need to convert it to a node-set using exsl:node-set.

The more portable approach is to use the document() function (many processors allow you to write a URIResolver or XmlResolver that intercepts the request and returns say a DOM Document node as the result), or to pass the document in as a stylesheet parameter.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
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