Hi,
Hi,Can somebody help me with the following issue: I want to
replace my
blank rows with the values from the preceeding non-blank
rows. A blank row
is a row containing all empty cells.
My xml is as follows:
<row>
<column name="firstname">K1</column>
<column name="lastname">L1</column
</row>
<row> --> Empty Row contains empty cells
<column name="firstname></column> --> These should be replaced by
<column name="lastname></column> --> "K1" and "L1" from prev. row.
</row>
<row> --> Empty row again.
<column name="firstname></column>
<column name="lastname></column>
</row>
....
Your XML is not well-formed. Anyhow, how about
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:param name="fillCell" select="'N'"/>
<xsl:template match="row[not(column/text())]">
<xsl:copy>
<xsl:choose>
<xsl:when test="$fillCell = 'Y'">
<xsl:for-each select="column">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:value-of
select="parent::row/preceding-sibling::row[column/text()][1]/column[(_at_)name
= current()/@name]" />
</xsl:copy>
</xsl:for-each>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="@*|node()"/>
</xsl:otherwise>
</xsl:choose>
</xsl:copy>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
You didn't specify it the source could look like this
<row>
<column name="firstname">XXX</column>
<column name="lastname"></column>
</row>
<row>
<column name="firstname"></column>
<column name="lastname"></column>
</row>
So the above doesn't work with it. It's easy to modify to handle it, thought.
Cheers,
Jarno - Anne Clark: Sleeper In Metropolis 3000 (Club)
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list