At 2009-10-27 11:00 -0700, Mark Wilson wrote:
In an XSLT-FO style sheet, I have created groups similar to the one
shown at the end of this email.
Fine.
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>).
What you are missing is that your current node at the beginning of
<xsl:for-each-group> is at the first member of the group selected.
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)">
At this point, <xsl:value-of select="CatalogNumber"/> will give you
"4" because your current node is the first <Item> of the group of
<Item> elements.
In the classroom I tell students to read the instruction as if it were written:
<xsl:for-the-first-member-of-each-group .....
... and then to work with the other members of the group using current-group().
<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.)
-->
You put the processing of the catalogue number before the
<xsl:for-each>, not inside.
<xsl:apply-templates select="CatalogNumber" mode="do"/>
Move that line up before the <xsl:for-each/>
<xsl:apply-templates select="Title" mode="do"/>
</xsl:for-each>
</fo:block>
</xsl:for-each-group>
I hope this helps.
. . . . . . . . . . . . . . Ken
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