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RE: [xsl] XSL Template Match using z:row attributes
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Subject: RE: [xsl] XSL Template Match using z:row attributes From: "Michael McCarty" <mccarty@xxxxxxxxxx> Date: Wed, 12 Dec 2001 20:22:26 -0800 |
Ok, you might need to ignore my previous reply. I've been playing with the XSL and (with the help of that Xselerator tool) I think I understand better, but not completly. My problem is that I'm getting duplicate output. The XML (trimmed) is: <?xml-stylesheet type="text/xsl" href="CategoryTree2.xsl" ?> <xml xmlns:rs='urn:schemas-microsoft-com:rowset' xmlns:z='#RowsetSchema'> <rs:data> <z:row LEVEL='1' CATNAME='Level 1' CATID='1' PARENTID='0' /> <z:row LEVEL='2' CATNAME='Level 2' CATID='3' PARENTID='1' /> <z:row LEVEL='2' CATNAME='Level 2' CATID='4' PARENTID='1' /> <z:row LEVEL='3' CATNAME='Level 3' CATID='5' PARENTID='4' /> <z:row LEVEL='3' CATNAME='Level 3' CATID='6' PARENTID='4' /> <z:row LEVEL='1' CATNAME='Level 1' CATID='21' PARENTID='0' /> <z:row LEVEL='2' CATNAME='Level 2' CATID='22' PARENTID='21' /> </rs:data> </xml> The XSL is: <?xml version='1.0'?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:rs='urn:schemas-microsoft-com:rowset' xmlns:z='#RowsetSchema' version="1.0"> <xsl:template match="/"> <HTML> <BODY topmargin="0" leftmargin="0"> <xsl:apply-templates select="/xml/rs:data"/> </BODY> </HTML> </xsl:template> <xsl:template match="z:row"> <div class="clsItem"> CatId_<xsl:value-of select="@CATID"/> - <b><xsl:value-of select="@CATNAME"/></b> - Parent_<xsl:value-of select="@PARENTID"/> <div id="container"> <xsl:apply-templates select="../z:row[(@LEVEL=current()/@LEVEL + 1) and (@PARENTID=current()/@CATID)]"/> </div> </div> </xsl:template> </xsl:stylesheet> Here's what I get: CatId_1 - Level 1 - Parent_0 CatId_3 - Level 2 - Parent_1 CatId_4 - Level 2 - Parent_1 CatId_5 - Level 3 - Parent_4 CatId_6 - Level 3 - Parent_4 CatId_3 - Level 2 - Parent_1 <!-- duplicate --> CatId_4 - Level 2 - Parent_1 <!-- duplicate --> CatId_5 - Level 3 - Parent_4 <!-- duplicate --> CatId_6 - Level 3 - Parent_4 <!-- duplicate --> CatId_5 - Level 3 - Parent_4 <!-- duplicate --> CatId_6 - Level 3 - Parent_4 <!-- duplicate --> CatId_21 - Level 1 - Parent_0 CatId_22 - Level 2 - Parent_21 CatId_22 - Level 2 - Parent_21 <!-- duplicate --> I think I'm going crazy learning this... =:-) As always, your help is greatly appreciated. Michael =:-) ---- TSchutzerWeissmann@xxxxxxxxxxxxxxxx wrote: > Michael wrote: > > >What I am getting is this: > > > >+ Product 1 > >+ Product 2 > > > >It appears that I get the LEVEL='1', but not the remaining. I wonder > >if there's a way around 'hard coding' the LEVEL's in case I have more > >than 4 or 5..?? I know it would be easier if my XML was is a different > >format, but I'm forced to use the MS ADO export format. Thank you, > in > >advance, for your help. > > Hi Michael, > > the problem here is something like this: your input isn't nested, and > you > want to nest it according to a LEVEL attribute. > There are three problems with your approach. > > 1. You're doing the same thing in 5 different templates. Try recursion > instead, then you don't need to worry about how many levels you have. > > > 2. The location paths in your <xsl:apply-templates>. > You use paths fine when you select attributes, so I assume you're > using the > full path (xml/rs:data/z:row) here because you want to get up out of > your > current context (a qualified z:row). > To make your paths work, either use /xml/rs:data/z:row - the "/" will > get > you back to the document node, or use ../z:row[@LEVEL=2]- in longhand > that's > parent::node()/z:row[@LEVEL=2]. It's just like file directories. > > BUT: it still won't work... > 3. Matching children with parents. > To get the nesting right you have to put the right children with the > right > parents. In your input you have CATEGORYID and PARENTCATEGORYID. A > revised > path, still using your approach would be > "../z:row[(@LEVEL=2) and (@PARENTCATEGORYID = > current()/@CATEGORYID)" > or even > "../z:row[(@LEVEL=current()/@LEVEL + 1) and (@PARENTCATEGORYID > = > current()/@CATEGORYID)" > You need current() not . because inside a predicate expression . is > the node > being evaluated, not the node matched by the current template. > > Only recurse... > Tom S-W > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > __________________________________________________ FREE voicemail, email, and fax...all in one place. Sign Up Now! http://www.onebox.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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