In XSLT 2.0 you can do
<xsl:for-each select="count(INITIALVALUES/INITIALVALUE)+1 to 5">
<tr>...</tr>
</xsl:for-each>
In 1.0 the usual approach is to select the first N nodes of some dummy
node-set:
<xsl:for-each select="(//node())[position() <= (5 -
count(INITIALVALUES/INITIALVALUE))]"
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Steve W [mailto:lsl(_at_)btconnect(_dot_)com]
Sent: 12 April 2005 07:58
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Having same number of rows in table
irrespective of number of data rows
I want to create a table that has a set number of rows in it.
The XML has
data in it to be shown in the table - in simple terms one
element of the XML
for each row of the table. The XML will only have an element
if there is
data, so if I have 3 data elements but I want 5 rows in the
table there
will be 2 'missing' rows and I want to output these 2 rows
with some set
html in it.
My template looks like this :
<table cellpadding="0" cellspacing="0" border="0" width="100%">
<xsl:for-each select="INITIALVALUES/INITIALVALUE">
<tr>
<td>
<!-- some html .... -->
</td>
</tr>
</xsl:for-each>
<!-- add 'blank' rows to give constant number of rows in table -->
</table>
Thanks
Steve
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