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Re: [xsl] Newbie document() question


Subject: Re: [xsl] Newbie document() question
From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>
Date: Fri, 12 Jan 2001 11:55:39 -0500

At 01/01/12 08:29 -0800, Edmund Mitchell wrote:
I've tried

<xsl:variable name='otherDoc' select='document(MyXml.xml)'/>
<xsl:template match='/'>
<xsl:copy-of select='$otherDoc'/>...

as well as just

<xsl:template match='/'>
<xsl:copy-of select='document(MyXml.xml)'/>...

but neither produced output, so I can't even seem to get off the ground with
this, let alone do any combining.

Don't kick yourself for this, Edmund, but you are asking for the document named by the value of the child element named MyXml.xml.


Try: select='document("MyXml.xml")'

This will supply the string of your filename to the function.

Note that you are working with a relative URI value here, so the presence and value of a second argument is going to be significant to where you want to find the file. Check Module 8 of your class notes (readers of our book will also find the details in chapter 8).

I hope this helps.

...................... Ken

--
G. Ken Holman                      mailto:gkholman@xxxxxxxxxxxxxxxxxxxx
Crane Softwrights Ltd.               http://www.CraneSoftwrights.com/s/
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Book:   Practical Transformation Using XSLT and XPath ISBN1-894049-05-5
Article: What is XSLT? http://www.xml.com/pub/2000/08/holman/index.html
Next public instructor-led training:             2001-01-27,2001-02-21,
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