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Re: [xsl] table formating

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Do a preprocessing pass to add an @index attribute to all cells.

Identity template plus

<xsl:template match="row">
<xsl:for-each-group select="cell" group-starting-with="cell[@index]">
<xsl:for-each select="current-group()">
<cell index="{current-group()[1]/@index + position() - 1}">
<xsl:copy-of select="child::node()"/>
</
</
</
</

Michael Kay
Saxonica

On 26/01/2011 13:50, Szabo, Patrick (LNG-VIE) wrote:

Hi,

Again This have to ask you people for help because I'm stuck.

This is an abstract of what my input looks like:

<row>
     <cell>
      <absatz>Sozialversicherung (1.650,00 * 18,07 %)</absatz>
     </cell>
     <cell index="3">  </cell>
     <cell>  </cell>
     <cell index="6">
      <absatz>298.16</absatz>
     </cell>
</row>

So the index of a<cell>  says what its position in a<row>  should be.
My desired output is:

<row>
     <cell>
      <absatz>Sozialversicherung (1.650,00 * 18,07 %)</absatz>
     </cell>
     <cell></cell>
     <cell index="3">  </cell>
     <cell>  </cell>
     <cell></cell>
     <cell index="6">
      <absatz>298.16</absatz>
     </cell>
</row>

NOTE: It's not sure that there are always 2 cells with @index. Could be
more or less too.

I've tried this:
<xsl:template match="cell[@index]">
   <xsl:variable name="prec_cells"
select="xs:integer(count(preceding-sibling::cell[not(@index)]) +
sum(preceding-sibling::cell/@index))"/>
		
	<xsl:for-each select="1 to xs:integer((@index - 1) -
$prec_cells)">
		<entry>
			<absatz/>
		</entry>
	</xsl:for-each>
		<entry>
			<xsl:apply-templates/>
		</entry>
</xsl:template>

Unfortunately this only works for 1<cell> with @index per<row>.

Can anyone tell me how to solve that ?!

Kind regards

. . . . . . . . . . . . . . . . . . . . . . . . . .
Patrick Szabo
  XSLT-Entwickler
LexisNexis
Marxergasse 25, 1030 Wien

mailto:patrick.szabo@xxxxxxxxxxxxx
Tel.: +43 (1) 534 52 - 1573
Fax: +43 (1) 534 52 - 146

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