Subject: Re: [xsl] recursive addition From: "M. David Peterson" <m.david@xxxxxxxxxx> Date: Wed, 14 Jan 2004 16:56:23 -0700 |
If all the folders within a root folder have the same 'id' (what would be the point of having the 'id' attribute on the child folder elements?) you could do this: <xsl:value-of select="sum(//folder[@id = 'dir0']/@files)"/> If they dont then this would do the trick: <xsl:value-of select="sum(//folder[@id = 'dir0']/@files) + sum(//folder[@id = 'dir0']//folder/@files)"/> Im even more confused from your explanation than I was the first time around but I think one of these will help you get to where you want to go. Best of luck, M. ----- Original Message ----- From: <annirack@xxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent: Wednesday, January 14, 2004 3:58 PM Subject: Re: [xsl] recursive addition > > The semantics are most definitely unclear but im assuming that you > > know what > > you want to gain out of adding dir2 and dir3. > > I apparently didn't explain myself well at all. Sorry about that, here's another attempt: > > I have an XML file that represents a file system, and has a flat file structure. > All folder nodes are siblings and all file nodes are children of a folder node. To represent a folder being within another folder, a file node can be a link to a folder node. > > Each folder node has an attribute that indicates how many files it contains. > > I want to know how many files are in all the subfolders of a given folder. > > I don't know if this is even possible in xsl 1.0 > > If it were not a flat file structure, it would seem easier... since I could select all subfolders using "root/folder[@id="some id"]//folder" and some form of recursive algorithm could probably handle the rest. > > --Brendan > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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