Re: [xsl] a sequence of regex-group()

["G. Ken Holman g.ken.holman@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

At 2015-10-28 15:20 +0000, Geert Bormans geert@xxxxxxxxxxxxxxxxxxx wrote:
> Hi all,
>
> sanity check
>
> If I wanted a sequence of all the regex-group() values available in 
> a matching-substring...
>
> (This is inside a function and the actual regex $re is passed in as 
> a param, so number of item() is unknown)
>
> I expected this to be somewhat more straightforward than this, so I check
>
>                 <xsl:variable name="group-count" 
> select="string-length(replace($re, '[^\(]+', ''))"/>
>                 <xsl:analyze-string select="$str" regex="{$re}">
>                     <xsl:matching-substring>
>                         <xsl:variable name="re-groups" select="for 
> $i in 1 to $group-count return regex-group($i)" as="item()*"/>
>
>
> Any thoughts?

Would it not be easier to somehow deal with the current context 
list?  Remember that analyze-string sets position() and last() for a 
sequence of matched and non-matched strings.

T:\>type temp.xsl
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
  xmlns:xs="http://www.w3.org/2001/XMLSchema";
  exclude-result-prefixes="xs"
  version="2.0">

<xsl:template match="/">
  <xsl:text>&#xa;</xsl:text>
  <xsl:analyze-string select="'a b c d e'" regex="\w">
    <xsl:matching-substring>
      <xsl:value-of select=".,position(),last(),'&#xa;'"/>
    </xsl:matching-substring>
    <xsl:non-matching-substring>
      <xsl:value-of select=".,position(),last(),'&#xa;'"/>
    </xsl:non-matching-substring>
  </xsl:analyze-string>
</xsl:template>

</xsl:stylesheet>
T:\>xslt2 temp.xsl temp.xsl
<?xml version="1.0" encoding="UTF-8"?>
a 1 9
  2 9
b 3 9
  4 9
c 5 9
  6 9
d 7 9
  8 9
e 9 9

I haven't looked at it in detail, but could you not determine the 
group count from the size of the context list?  Perhaps not, if you 
don't know if the first or last is matching or non-matching.

I hope this becomes helpful.  I am on a conference call and sneaking 
in this response.

. . . . . . . . Ken


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