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Re: [xsl] Generate separate elements, not just attribute values

2017-03-23 15:48:16
Thanks Martin and Michael for the help and the information. Greatly appreciated!

When I saw your responses, I got a bit of a laugh at how bad my original .xsl 
was.

-Charles

-----Original Message-----
From: Martin Honnen martin(_dot_)honnen(_at_)gmx(_dot_)de 
[mailto:xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com] 
Sent: Thursday, March 23, 2017 4:02 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Generate separate elements, not just attribute values

On 23.03.2017 19:23, Charles O'Connor coconnor(_at_)ariessys(_dot_)com wrote:
Hi all,

Extreme novice here, so I appreciate your help.

Using XSLT 2.0 (explained later) and Saxon9 HE.

I have JATS 1.1 (archiving) input:

. . .

<contrib-group>
      <contrib contrib-type="author">
              <name>
                      <surname>Franzke</surname>
                      <given-names>Christian L. E.</given-names>
              </name>
      </contrib>
      <aff id="aff2">Meteorological Institute and Center for Earth System 
Research and Sustainability (CEN), <institution>University of 
Hamburg</institution>, <country>Germany</country></aff>
      <aff id="aff3">Department of Cheddar, <institution>University of 
Curds and Whey</institution>, <country>Land of Cheese</country></aff> 
</contrib-group>

. . .

To make it easier for our engineers to recognize relationships between 
<contrib>s and <aff>s in cases where they are related through nesting in a 
<contrib-group>, not <xref>s, I wrote a small .xsl:

<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="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="//contrib[not(xref/@ref-type='aff')]">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
            <xsl:element name="xref">
                <xsl:attribute name="ref-type">aff</xsl:attribute>
                <xsl:attribute name="rid">
                    <xsl:value-of select="parent::contrib-group/aff/@id"/>
                </xsl:attribute>
            </xsl:element>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

Which gives as output

. . .

<contrib-group>
      <contrib contrib-type="author">
              <name>
                      <surname>Franzke</surname>
                      <given-names>Christian L. E.</given-names>
              </name>
              <xref ref-type="aff" rid="aff2 aff3"/>
      </contrib>
      <aff id="aff2">Meteorological Institute and Center for Earth System 
Research and Sustainability (CEN), <institution>University of 
Hamburg</institution>, <country>Germany</country></aff>
      <aff id="aff3">Department of Cheddar, <institution>University of 
Curds and Whey</institution>, <country>Land of Cheese</country></aff> 
</contrib-group>

. . .

But in cases of more than one <aff> in a <contrib-group>, instead of

      <xref ref-type="aff" rid="aff2 aff3"/>

it turns out the engineers really want

      <xref ref-type="aff" rid="aff2"/><xref ref-type="aff" rid="aff3"/>

Do I need to use a "for-each" to do this? I was beginning to think I was 
halfway clever when I figured out I didn't need a "for-each" to get the 
initial .xsl to work, but maybe not so clever after all.

You can use for-each select="parent::contrib-group/aff/@id" or
   <xsl:apply-templates select="parent::contrib-group/aff/@id" mode="ref"/>

and then

<xsl:template match="aff/@id" mode="ref">
   <xref ref-type="aff" rif="."/>

Incidental question: I have the version as 2.0 because, well, that was 
the version on the identity template I copied from wherever. I didn't 
see any reason for it to be 2.0, however, and 1.0 would be easier 
because you can run 1.0 in .NET without additional software. But, when 
I changed the version to 1.0, it ran fine but only gave the first @rid 
value, i.e.,

<xref ref-type="aff" rid="aff2"/>

Why?

With XSLT 1.0, xsl:value-of select="parent::contrib-group/aff/@id" 
outputs only the value of the first selected node because it is defined that 
way in XSLT 1.0 (https://www.w3.org/TR/xslt#value-of says to convert the 
node-set to a string and https://www.w3.org/TR/xpath/#function-string defines 
the string value of a node-set to be the string value of the first node). And 
XSLT 2.0 processors supporting backwards compatibility (like Saxon 9) are 
required to emulate that XSLT/XPath 1.0 behaviour 
(https://www.w3.org/TR/xslt20/#backwards,
https://www.w3.org/TR/xslt20/#value-of "Special rules apply when backwards 
compatible behavior is enabled for the instruction. If no separator attribute 
is present, and if the select attribute is present, then all items in the 
atomized result sequence other than the first are ignored.").
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