convert XSLT 2.0 to 1.0 - xslt

Question

I have the following solution in XSLT 2.0
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/rows">
<xsl:copy>
<xsl:for-each-group select="row" group-starting-with="row[item='****************']">
<xsl:variable name="title-item" select="preceding-sibling::row[1]/item" />
<xsl:for-each select="current-group()[starts-with(item[1], '#')]">
<row>
<xsl:copy-of select="$title-item, *"/>
</row>
</xsl:for-each>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
as for-each-group is not supported in XSLT 1.0
I want to convert it into XSLT 1.0 version.

Answer 1

Declare e.g.
<xsl:key name="group" match="row[not(item = '****************')]" use="generate-id(preceding-sibling::row[item='****************'][1])"/>
as a top-level element, then instead of <xsl:for-each-group select="row" group-starting-with="row[item='****************']"> try
<xsl:for-each select="row[item='****************']">
<xsl:variable name="group" select="key('group', generate-id())"/>
<xsl:variable name="title-item" select="preceding-sibling::row[1]/item" />
<xsl:for-each select="$group[starts-with(item[1], '#')]">
<row>
<xsl:copy-of select="$title-item | *"/>
</row>
</xsl:for-each>
</xsl:for-each>
But we really need to see some input sample structure plus the wanted result. And I might be overlooking some XSLT/XPath 2 only stuff I have used despite trying to write XSLT 1.