Determining files in a directory using xslt - xslt

Question

I am replacing the Distribution section in my many metadata records to include optional references to other non-xml files. Each xml metadata record is in a separate folder, and in each such folder there may or may not be a PDF document and perhaps an Excel file. These three files have independent names. I basically need a directory list, which I can access and manipulate.
The following is a subset of what I have been attempting, although as it stands it does compile using Kernow-Saxon 9PE. The first part seems to construct the correct path to the desired subdirectory, albeit with a “file:/” prefix.
My problems are:
Do I need to go to the effort I am making to determine the path? (It seems “un-xsl-ish”). Kernow easily finds the metadata.xml file in each subdirectory.
I don’t think I have the “for-each” statement correct. In versions where it does execute, I temporarily limit “select=.” as “select=*.xml” (and it only does the metadata.xml files).
Without the override mentioned, I get “Variable filename has not been declared (or its declaration is not in scope)”
<xsl:strip-space elements="*" />
<xsl:template match="gmd:distributionInfo" >
<gmd:distributionInfo>
<gmd:distributorTransferOptions>
<!-- construct path to directory containing xml metadata and non-xml files -->
<xsl:variable name="pathParts" select="tokenize(base-uri(),'/') " />
<!-- remove filename.extension, reconstruct path -->
<xsl:variable name="directory" select="string-join((remove($pathParts,count($pathParts))),'/')" />
<gmd:dir>
<gco:CharacterString>
<xsl:value-of select = "$directory" />
</gco:CharacterString>
</gmd:dir>
<!-- loop thru all files to obtain filenames -->
<xsl:for-each select="for $filename in collection(concat($directory, select='*.*')) return $filename " >
<!-- temporary? override to enable xslt compilation -->
<xsl:variable name="filename" select= "base-uri()" />
<gmd:name>
<gco:CharacterString>
<xsl:value-of select= "$filename" />
</gco:CharacterString>
</gmd:name>
</xsl:for-each>
</gmd:distributorTransferOptions>
</gmd:distributionInfo>
</xsl:template>
<xsl:template match="*" >
<xsl:copy>
<xsl:copy-of select="#*" />
<xsl:apply-templates />
</xsl:copy>
</xsl:template>
</xsl:transform>
A dummy metadata record would as attached.
<?xml version="1.0" encoding="UTF-8"?>
<gmd:MD_Metadata xmlns:gco="http://www.isotc211.org/2005/gco" xmlns:gmd="http://www.isotc211.org/2005/gmd">
<gmd:fileIdentifier>
<gco:CharacterString>56400C76-E5E3-44D7-904C-90B97858F7CE</gco:CharacterString>
</gmd:fileIdentifier>
<gmd:identificationInfo>
<gmd:MD_DataIdentification>
</gmd:MD_DataIdentification>
</gmd:identificationInfo>
<gmd:distributionInfo>
<gmd:MD_Distribution>
</gmd:MD_Distribution>
</gmd:distributionInfo>
</gmd:MD_Metadata>

Answer 1

You cannot access the defined $filename outside of your xpath for $i in x statement.
Try making the following changes to your code:
<xsl:for-each select="for $filename in collection(concat($directory, select='*.*')) return $filename " >
<gmd:name>
<gco:CharacterString>
<xsl:value-of select= "base-uri(.)" />
</gco:CharacterString>
</gmd:name>
</xsl:for-each>
The select statement in the for-each should return a document node for each item in your collection, base-uri(.) will give you the filename for each document node.
Also if you only have xml files in your collection I would use select='*.xml' because if there were other files in your directory that statement would fail.