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(_dot_)szabo(_at_)lexisnexis(_dot_)at
Tel.: +43 (1) 534 52 - 1573
Fax: +43 (1) 534 52 - 146
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