I just remembered that isn't the only way ... you can also use base-uri():
t:\ftemp>xslt2 mark.xml mark.xsl
Filename: file:/t:/ftemp/mark.xml
Alternate: file:/t:/ftemp/mark.xml
t:\ftemp>type mark.xsl
<?xml version="1.0" encoding="US-ASCII"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0">
<xsl:output method="text"/>
<xsl:template match="/">
Filename: <xsl:value-of select="document-uri(/)"/>
Alternate: <xsl:value-of select="base-uri(/)"/>
</xsl:template>
</xsl:stylesheet>
t:\ftemp>
At 2009-08-20 21:22 -0400, I wrote:
At 2009-08-20 18:18 -0700, Mark Wilson wrote:
I've poked around, but suspect I am searching for the wrong terms
in the index to Michael's XML/XPath book. I found one reference on
the web that said what I want to do could not be done with XSLT 1.0.
You can do it for the stylesheet in XSLT 1.0 but not for the data file.
However, I am using XSLT 2.0.
Not a problem.
I need to know the file name of the file my XSLT style sheet is
acting upon so that I can write it out as part of the tile of an
XSL-FO output PDF document. Does XSLT 2.0 have a way of getting the file name?
The document-uri() function is what you need.
I hope the example below helps.
. . . . . . . . Ken
t:\ftemp>xslt2 mark.xml mark.xsl
Filename: file:/t:/ftemp/mark.xml
t:\ftemp>type mark.xsl
<?xml version="1.0" encoding="US-ASCII"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0">
<xsl:output method="text"/>
<xsl:template match="/">
Filename: <xsl:value-of select="document-uri(/)"/>
</xsl:template>
</xsl:stylesheet>
t:\ftemp>
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