You may try the following XSL --
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:msxsl="urn:schemas-microsoft-com:xslt">
<xsl:output method="xml" version="1.0"
encoding="UTF-8" indent="yes"/>
<xsl:template match="/nr">
<xsl:variable name="result">
<xsl:for-each select="featured/movie">
<a/>
</xsl:for-each>
<xsl:for-each select="also_new/movie">
<a/>
</xsl:for-each>
</xsl:variable>
<xsl:for-each select="msxsl:node-set($result)/a">
<xsl:value-of select="position()"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Regards,
Mukul
--- Cynthia DeLaria <cdelaria(_at_)quris(_dot_)com> wrote:
Good Day,
I have searched the xsl list unsuccessfully for an
answer to this
question, although I'm sure something like this has
been addressed. I
think I'm just not sure what to search on to find
it.
I have the following xml snippet:
<nr>
<featured>
<movie title="Charlie's Angles: Full
Throttle">Description</movie>
<movie title="28 Days
Later">Description</movie>
<movie title="The Santa Clause
2">Description</movie>
</featured>
<also_new>
<movie title="Northfork" />
<movie title="Rudy: The Rudy Giuliani Story"
/>
<movie title="Russian Ark" />
</also_new>
</nr>
Basically, in the fully-flushed out version of the
XML, the <featured>
movies have images and full descriptions, while the
<also_new> movies
have only a title and rating. What I need to do is
find a way to find
the position of each <movie> node relative to the
root, as I need to
create a "print the new releases" page that lists
all new releases in
two columns. This is what I tried, but it gives me
the position based on
the parent (i.e. <also_new> or <featured>) not
relative to the root.
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html" encoding="ISO-8859-1" />
<xsl:template match="/nr">
<html>
<head>
<title>New Releases E-Newsletter</title>
<head>
<body bgcolor="#ffffff" text="#000000"
link="#023f7e" alink="#ff0000"
vlink="#023f7e">
Just print out this list.<br />
<table width="100%">
<xsl:variable name="thisMany"><xsl:value-of
select="(count(//*[name()='movie']) div 2)"
/></xsl:variable>
<tr>
<td valign="top" width="50%" class="body2">
<br />
<xsl:for-each select="//movie[position() <=
number($thisMany)]">
<b><i><xsl:value-of select="./@title"
/></i></b><br />
</xsl:for-each>
<br />
</td>
<td valign="top" width="50%" class="body2">
<br />
<xsl:for-each
select="//*[name()='movie'][position() >
number($thisMany)]">
<b><i><xsl:value-of select="./@title"
/></i></b><br />
</xsl:for-each>
<br />
</td>
</tr>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
Is it possible to get the position relative to the
root node? It seems
like this should be very simple, but all of the
things I have tried have
not worked to produce the intended outcome.
Thank you!
Cynthia
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