Subject: [xsl] convert nested elements, ID's and classes to CSS declaration From: xslt user <xsltacct@xxxxxxxxx> Date: Wed, 21 Mar 2007 09:36:30 -0700 (PDT) |
Hello, I searched for this all over and couldn't find a solution. I want to take the following XML: <root> <file> <element/> </file> <file id="a"> <element id="a"/> </file> <file class="b"> <element class="b"/> </file> <file id="c" class="d" not="1" this="2"> <element id="c" class="d" not="1" this="2"/> </file> </root> #### and output something this #### root root file root file element root file #a root file #a element #a root file .b root file .b element .b root file #c .d root file #c .d element #c .d #### ignoring any attributes that are not "id" or "class" and containing all ancestor elements and id's and classes no matter what the element name is or how many levels deep. I'd like to use this for auto generating CSS declarations from XHTML snipits, I can deal with removing redundant rows manually. Below is the closest I've gotten which only strips the XML to the elements and attributes I'm interested in, but doesn't format it or list ancestor elements, id's and classes (I'm a beginner at XSLT) thanks! #### <xsl:template match="node()|@id|@class"> <xsl:copy> <xsl:apply-templates select="node()|@id|@class"/> </xsl:copy> </xsl:template> ____________________________________________________________________________________ We won't tell. Get more on shows you hate to love (and love to hate): Yahoo! TV's Guilty Pleasures list. http://tv.yahoo.com/collections/265
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] document() and key() boun, Michael Kay | Thread | [xsl] Question Regarding the displa, Tim Golen |
RE: [xsl] xpath / multiple context , Angela Williams | Date | [xsl] Question Regarding the displa, Tim Golen |
Month |