Subject: Re: [xsl] how to rearrange nodes based on a dependency graph? From: Gunther Schadow <gunther@xxxxxxxxxxxxxxxxxxxxxx> Date: Thu, 20 Dec 2001 17:52:35 -0500 |
I still can't see what this would do. This isn't about simply filtering out nodes with dependencies, it is filtering out those nodes that had already been processed because of the dependency graph walk. The only way I can see to do this is to somewhere keep a list of the processed nodes, but how can I do that with XSL? (I guess I can't). The only alternative would be to decide for every node if it would have been in the dependency graph of any prior node. But that would mean that all dependency graphs need to be always chased for every node, that's kind of mind-boggling.
Or even <xsl:key name='fragkey' match='frag' use='@id' /> <xsl:template match="frag[@requires]"> <xsl:for-each select="@requires"> <xsl:for-each select="key('fragkey',.)[not(@requires)]"> <xsl:copy-of select="." /> </xsl:for-each> </xsl:for-each> <xsl:copy-of select="." /> </xsl:template> <xsl:template match="frag" />
Hey it's christmas. I don't use ids and idrefs. If this <xsl:for-each select="@requires"> returns all of the idrefs then this might work. But if it doesn't then mine is a stella and a bacardi and coke hick!
Ciao Chris
XML/XSL Portal http://www.bayes.co.uk/xml
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
-- Gunther Schadow, M.D., Ph.D. gschadow@xxxxxxxxxxxxxxx Medical Information Scientist Regenstrief Institute for Health Care Adjunct Assistant Professor Indiana University School of Medicine tel:1(317)630-7960 http://aurora.regenstrief.org
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] how to rearrange nodes ba, Chris Bayes | Thread | Re: [xsl] how to rearrange nodes ba, Wendell Piez |
RE: [xsl] how to rearrange nodes ba, Chris Bayes | Date | Re: [xsl] how to rearrange nodes ba, Wendell Piez |
Month |