On 05/06/2025 20:49, rick@xxxxxxxxxxxxxx wrote:
>
> Here is my XML:
>
> <?xml version="1.0" encoding="UTF-8"?>
>
> <root>
>
> B B B  <record>
>
> B B B B B B B  <col1>PG</col1>
>
> B B B  </record>
>
> B B B  <record>
>
> B B B B B B B  <col1>1</col1>
>
> B B B  </record>
>
> B B B  <record>
>
> B B B B B B B  <col1>2</col1>
>
> B B B  </record>
>
> B B B  <record>
>
> B B B B B B B  <col1>PG</col1>
>
> B B B  </record>
>
> B B B  <record>
>
> B B B B B B B  <col1>3</col1>
>
> B B B  </record>
>
> B B B  <record>
>
> B B B B B B B  <col1>4</col1>
>
> B B B  </record>
>
> </root>
>
> I want to group all of the <record> elements, starting with
> col1[starts-with(.,bPGb)]. I am expecting 2 groups of 3 record
> elements each.
>

A variant of your code with e.g.

 B B B  <xsl:template match="root">

 B B B B B B B  <xsl:copy>


 B B B B B B B B B B B  <xsl:for-each-group select="record"
group-starting-with="record[col1[.='PG']]">

 B B B B B B B B B B B B B B B  <group></group>

 B B B B B B B B B B B  </xsl:for-each-group>

 B B B B B B B  </xsl:copy>

 B B B  </xsl:template>

would do that. Remember, the group-starting-with is a pattern the
element to start a group has to match.