[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] how to replace an XML tag with a result tree and parse
Hi i have some difficulties with the following problem: In my xml i invented a tag like this: <toc level="2"/> to indicate where to put a table of contents, so i dont want to put ToC automatically. only if the author wants to show it. in my xsl i am trying to do this: transform the toc tag into xml again which has the same DTD as the source document. if dtd is something like docbook i would like to replace <toc level=2"/> with <itemizedlist> <listitem>Chapter 1</listitem> <listitem> Chapter 2</listitem> ... </itemizedlist> i put this result tree in a varaible like <xsl:variable name="mytoc"> <itemizedlist> <listitem>Chapter 1</listitem> <listitem> Chapter 2</listitem> ... </itemizedlist> </xsl:varaible> (of course it looks more comlicated, just simplifying things) then i try to <xsl:apply-templates select="$mytoc"/> it works, but the tree in $mytoc has no knowledge about the input tree in xml. so if my <itemizedlist> needs to know to which bookset it belongs to do things like this <a> <xsl:attribute name="class"> <xsl:value-of select="ancestor-or-self::set/@id"/> </xsl:attribute> ... </a> then i am in trouble, because the tree in $mytoc doesnt belong to the main xml tree. So here is my question: How can i achieve that templates replace something in the input tree and the Stylesheet behaves as it was always there? thanks in advance janning -- Planwerk 6 /websolutions Herzogstraße 86 40215 Düsseldorf fon 0211-6015919 fax 0211-6015917 http://www.planwerk6.de XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|