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Home/ Questions/Q 1066719
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T20:03:30+00:00 2026-05-16T20:03:30+00:00

I need to analyze a few thousand XML documents to see if some of

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I need to analyze a few thousand XML documents to see if some of them contains a certain construct. The problem is that some of the documents doesn’t contain well formed XML.

The basic idea was to use fn:collection() and search inside nodes returned. But this only works if all documents in the collection are well formed.

Is it possible to do something similar but only parsing the well formed documents?

This is my XSLT, simplified, which works if all documents in $dir are well formed:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet 
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"
  xmlns:xs="http://www.w3.org/2001/XMLSchema">

  <xsl:output method="text"/>
  <xsl:variable name="dir" as="xs:string">file:/c:/path/to/files/</xsl:variable>
  <xsl:variable name="files" select="concat($dir, '?select=*.xml')" as="xs:string"/>

  <xsl:template match="/">
    <xsl:variable name="docs" select="collection($files)"/>
    <xsl:variable name="names" select="
      for $i in $docs return
        distinct-values($i//*[exists(@an-attribute-to-find)]/local-name())"/>
    <xsl:value-of select="distinct-values($names)" separator="&#x0a;"/>
  </xsl:template>

</xsl:stylesheet>

Would it be possible to do something like this without manually sorting out the non well formed documents before transformation starts? Maybe you have a better suggestion to a solution?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T20:03:30+00:00Added an answer on May 16, 2026 at 8:03 pm

    At present this is best done out of XSLT.

    It can be done in XSLT if you provide as an exrternal parameter (<xsl:param>) to the transformation a list of all filenames to be processed — then the transformation would use the standard XPath 2.0 function doc-available() and operate only on the document nodes returned by this function.

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