<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0" media-type="text/html"
indent="yes"
standalone="yes"
encoding="ISO-8859-1" omit-xml-declaration="yes"/>
<xsl:template match="/document">
<html xmlns="http://www.w3.org/1999/xhtml"; lang="it" xml:lang="it">
<head>
<title> <xsl:value-of select="@title"/> </title>
</head>
<body>
<h1>Some Header H1</h1>
<div id="header">
<a href="#" title="Home page" accesskey="1" tabindex="1">
Home page
</a>
</div>
<dd>
<ul>
<li>
<a href="#" accesskey="p" tabindex="2">Entry 1</a>
</li>
<li>
<a href="#" accesskey="t" tabindex="2">Entry 2</a>
</li>
<li>
<a href="#" accesskey="a" tabindex="3">Entry 3</a>
</li>
</dd>
<dd>
<ul>
<li>
<a href="#" accesskey="a" tabindex="4">Entry 4</a>
</li>
<li>
<a href="#" accesskey="b" tabindex="5">Entry 5</a>
</li>
<li>
<a href="#" accesskey="c" tabindex="6">Entry 6</a>
</li>
</dd>
<div id="content"><p><xsl:value-of select="content"/></p></div>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
This is complete trasformation...I read an XML document that contains
only one tag with content and attribute title. I get them and display
in right position. Links in my document are stati link, organized in
list, unordered list. If I use position i get the position number in
the current node list of the node that is currently being processed,
or rather get same position for all tabindex attribute.
I would want to generate value of tabindex attribute in such way that
can add or reorganize my link tags without re-organize all tabindex.
Regards,
Mulp.
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--