XSLT: analyze-string on text with children nodes

Question

I am trying to use analyze-string on elements that possibly contain children nodes. Here is a snippet of a source file: <year> 1975 music </year> <year> 1985<genre>rock</genre>music </year> <year> 2005 music </year>

And here is what I would like to get:

    <year>
        70's music
    </year>
    <year>
        80's<genre>rock</genre> music
    </year>

The following code correctly matches the 3 cases, but I am missing something to copy both the text and the possible children nodes under the year element:

<xsl:template match="year">
    <xsl:variable name="item" select="."/>
    <xsl:analyze-string select="." regex="19(\d)\d">
        <xsl:matching-substring>
            <xsl:variable name="decade">
                <xsl:value-of select="concat(regex-group(1),'0''s')"/>
            </xsl:variable>
            <year>
                <xsl:value-of select="concat($decade,regex-group(2))"/>
                <genre><xsl:value-of select="$item/genre"/>
                </genre>

            </year>
        </xsl:matching-substring>
        <xsl:non-matching-substring> </xsl:non-matching-substring>
    </xsl:analyze-string>
</xsl:template>

Here is what I get:

    <year>70's<genre/></year>
    <year>80's<genre>rock</genre></year>

Thanks in advance for any help!

Answer 1

It seems to suffice to match on the text child for the year pattern:

  <xsl:template match="year[not(matches(., $pattern))]"/>

  <xsl:template match="year/text()[matches(., $pattern)]">
      <xsl:value-of select="replace(., $pattern, '$10''s')"/>
  </xsl:template>

That gives

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="#all"
    version="3.0">

  <xsl:mode on-no-match="shallow-copy"/>

  <xsl:param name="pattern" as="xs:string">19(\d)\d</xsl:param>

  <xsl:output method="xml" indent="yes"/>

  <xsl:template match="year[not(matches(., $pattern))]"/>

  <xsl:template match="year/text()[matches(., $pattern)]">
      <xsl:value-of select="replace(., $pattern, '$10''s')"/>
  </xsl:template>

</xsl:stylesheet>

https://xsltfiddle.liberty-development.net/gWmuiJV gives the result

<root>
       <year>
            70's music
        </year>
        <year>
            80's<genre>rock</genre>music
        </year>

</root>

For an XSLT 2 compatible solution you would need to spell out the xsl:mode declaration used as the identity transformation template, example for that at http://xsltransform.hikmatu.com/jyH9rLQ.