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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] XML structure to HTML unordered list
 I'm struggling to figure out a good way to transform a XML structure into a HTML unordered list. So far the solution I've come up with looks like this: Sample index.xml file <?xml version="1.0" encoding="iso-8859-1"?>
<?xml-stylesheet type="text/xsl" href="index.xsl"?>
<root_element>
<item id="1">
desc 1
<item id="11">
desc 11
</item>
<item id="12">
desc 12
<item id="121">
desc 121
<item id="1211">
desc 1211
<item id="12111">
desc 122111
<item id="1211111">
desc 1211111
<item id="12111111">
desc 12111111
</item>
<item id="12111112">
desc 12111112
</item>
</item>
</item>
</item>
</item>
</item>
</item>
<item id="2">
desc 2
<item id="21">
desc 21
</item>
<item id="22">
desc 22
<item id="221">
desc 221
</item>
<item id="222">
desc 222
</item>
</item>
<item id="23">
desc 23
</item>
</item>
</root_element>And here's my index.xsl file <?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="html" omit-xml-declaration="yes"/> <xsl:template match="root_element">
<html>
<body>
<xsl:call-template name="unordered-list">
<xsl:with-param name="items" select="item"/>
</xsl:call-template>
</body>
</html>
</xsl:template> <xsl:template name="unordered-list">
<xsl:param name="items" select="/.."/>
<ul>
<xsl:for-each select="$items">
<li>
<xsl:value-of select="text()"/>
<xsl:if test="item">
<xsl:call-template name="unordered-list">
<xsl:with-param name="items" select="item"/>
</xsl:call-template>
</xsl:if>
</li>
</xsl:for-each>
</ul>
</xsl:template></xsl:stylesheet> The above produces what I want, and it seems quite solid. But is there perhaps another (better?) way to solve this task? I'm limited to XSLT 1.0. Sincerely, Thomas 
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