[xsl] XSL-FO group by problem

Subject: [xsl] XSL-FO group by problem
From: "Mark Wilson" <mark@xxxxxxxxxxxx>
Date: Tue, 27 Oct 2009 11:00:41 -0700
Hello all,
In an XSLT-FO style sheet, I have created groups similar to the one shown at the end of this email.
I have one template for formatting the <CatalogNumber> and another for formatting the remaining data. The <CatalogNumber> is always identical in a group, but the remaining information differs. I want a single copy of the <CatalogNumber> and one copy each of all of the remaining information from each <Item>as indicated in the output below (I can format everything, that's not my problem, I just can't get the single copy of the <CatalogNumber>).


Thanks,
Mark

Output:
4: New issues: Czech Republic, May/Jun 1993, p. 22; Letters to the editor, Mar/Apr 2002, p. 27; Joint issues, Nov/Dec 2002, p.18.


My style sheet code is:
<xsl:for-each-group select="Item" group-by="concat(Prefix, CatalogNumber, Range)">
<fo:block xsl:use-attribute-sets="base">
<xsl:for-each select="current-group()">


<!-- What do I put here so that I only get one copy of the Catalog Number?
(This code seems to go into an endless loop.)
-->
<xsl:apply-templates select="CatalogNumber" mode="do"/>
<xsl:apply-templates select="Title" mode="do"/>


               </xsl:for-each>
           </fo:block>
       </xsl:for-each-group>

Example of a group:
<Item>
<CatalogNumber>4</CatalogNumber>
<Title>New issues: Czech Republic</Title>
<IssueName>May/Jun</IssueName>
<Year>1993</Year>
<Page>22</Page>
</Item>
<Item>
<CatalogNumber>4</CatalogNumber>
<Title>Letters to the editor</Title>
<IssueName>Mar/Apr</IssueName>
<Year>2002</Year>
<Page>27</Page>
</Item>
<Item>
<CatalogNumber>4</CatalogNumber>
<Title>Joint issues</Title>
<IssueName>Nov/Dec</IssueName>
<Year>2002</Year>
<Page>18</Page>
</Item>


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