Subject: [xsl] XML structure to HTML unordered list From: Kaz <kaz@xxxxxxxxxxxxxxxxxxx> Date: Sun, 09 Sep 2007 12:49:14 +0200 |
<?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>
<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>
Sincerely, Thomas
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