Re: Finding Only Initial Following Siblings That Meet

In my case, I must start with the first instance of the matching phrase
anywhere in the source document (I'm pulling stuff that could be anywhere to a
specific location) and then only want to consider things that immediately
follow that specific <ph> element.

So unless I'm missing a subtlety of your solution, I don't think it would do
quite what I want because it's too inclusive.

Cheers,

E.
--
Eliot Kimber
http://contrext.com


o;?On 2/5/20, 5:07 PM, "Imsieke, Gerrit, le-tex gerrit.imsieke@xxxxxxxxx"
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

    Grouping should liberate you from looking ahead or behind. So instead of
    matching the first <ph outputclass="x">, you'd match <p> (or more
    generally '*[ph[@outputclass]]') and do the group-adjacent grouping for
    the child nodes, like this:

    <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
       version="3.0">
       <xsl:template match="*[ph[@outputclass]]">
         <xsl:copy>
           <xsl:apply-templates select="@*" mode="#current"/>
           <xsl:for-each-group select="node()"
             group-adjacent="string(self::ph/@outputclass)">
             <xsl:choose>
               <xsl:when test="current-grouping-key()">
                 <xsl:element name="{current-grouping-key()}">
                   <xsl:value-of select="current-group()"
                     separator=""/>
                 </xsl:element>
               </xsl:when>
               <xsl:otherwise>
                 <xsl:apply-templates select="current-group()"
                   mode="#current"/>
               </xsl:otherwise>
             </xsl:choose>
           </xsl:for-each-group>
         </xsl:copy>
       </xsl:template>

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

    This is not shorter in terms of lines of code than what you suggested.
    In terms of performance, it could be a bit more efficient than your
    solution, depending on the cost of identifying the first
    ph[@output-class] and its following siblings, compared to the cost of
    identifying a parent of ph[@output-class] and selecting its children.

    But as I wanted to say above, in terms of idiomatic XSLT 2+ purity, I'd
    always prefer a solution that doesn't look along the preceding/following
    axes, even when it is done just once for selecting the for-each-group
    population.

    Gerrit

    On 05.02.2020 23:29, Eliot Kimber ekimber@xxxxxxxxxxxx wrote:
    > In my XML I can have adjacent elements that should be processed as a
unit, where the adjacent elements all have the same value for a given
attribute. Other elements with the same attribute could be following siblings
but separated by other elements or text nodes, i.e.:
    >
    > <p>Text <ph outputclass="x">1</ph><ph outputclass="x">2</ph> more text
<ph outputclass="x">New sequence</ph></p>
    >
    > Where the rendered result should combine the first two <ph> elements but
not the third, i.e.:
    >
    > <p>Text <x>12</x> more text <x>New sequence</x></p>
    >
    > Processing is applied to the first element in the document with the
@outputclass value "x" and then I want to grab any immediately following
siblings with the same @outputclass value and no intervening text or element
nodes.
    >
    > My solution is to use for-each-group like so:
    >
    >      <xsl:variable name="this" as="element()" select="."/>
    >      <xsl:variable name="adjacent-sibs" as="element()+">
    >        <xsl:for-each-group select="($this,
$this/following-sibling::node())"
    >          group-adjacent="string(@outputclass)">
    >          <xsl:if test=". is $this">
    >            <xsl:sequence select="current-group()"/>
    >          </xsl:if>
    >        </xsl:for-each-group>
    >      </xsl:variable>
    >
    > Which works, but I'm thinking there must be a more compact way to do the
same selection, but the formulation is escaping me.
    >
    > Is there a more compact or more efficient way to make this selection of
only immediately-adjacent following siblings?
    >
    > Thanks,
    >
    > E.
    > --
    > Eliot Kimber
    > http://contrext.com