Re: [xsl] XSL resources - Flat to hierarchy - Common ancestors

["M. David Peterson" <m.david@xxxxxxxxxx>]

Hey Ben,

Ben Simkins wrote:
> Three questions in one mail:
> 1. Does anyone know of a public 'knowledgebase' type site where there
> are lots of examples of solutions to different problems, classified in a
> way which avoids spending hours trawling through threads?

Yep... Dave Pawson maintains such a site and can be found at 
http://www.dpawson.co.uk/xsl/index.html
> 2. Transforming 'flat' structures to hierarchies: 
> I've adapted a stylesheet found in this list, for creating the hierarchy
> of this:
> <Menus>
>   <Menu MenuId="58" MenuName="LeftMenu"/>
>   <Menu MenuId="1" MenuParentId="58" MenuName="Home"/>
>   <Menu MenuId="60" MenuParentId="1" MenuName="About us/>
>   ...etc
> </Menus>
>
> The xsl looks like this:
> <xsl:template match="/">
> 	<xsl:apply-templates select="Menus"/>
> </xsl:template>
>
> <xsl:template match="Menus">
>   <xsl:copy>
>     <xsl:apply-templates select="Menu[not(@MenuParentId)]">
>     </xsl:apply-templates>
>   </xsl:copy>
> </xsl:template>
>
> <xsl:template match="Menu">
>   <xsl:copy>
>   	<xsl:copy-of select="@*"/>
>   	<xsl:apply-templates select="../Menu[@MenuParentId =
> current()/@MenuId]"/> 
>   </xsl:copy>
> </xsl:template>
>
> Question: is there any way of also adding the numbering (<xsl:number
> level="multiple" format="1.1"/>) directly, or do I have to run the
> output through a second stylesheet?

Wrap the first apply-templates into a variable, use your processors node-set() 
function (see your processors documentation on how to implment this) to convert 
the result into a node-set and then apply-templates or for-each through the 
resulting node-set using <xsl:number/> to accomplish the task

<xsl:template match="/">
	<xsl:variable name="menus"
	<xsl:apply-templates select="Menus"/>
	</xsl:variable>
	<xsl:for-each select="exsl:node-set($menus)">
<!-- change the namespace and syntax for the node-set function above specific to 
the what the docs tell you for your processor - dont forget to declare the 
namespace in the xsl:stylesheet element! -->
		<xsl:number/>
		.. whatever else you want to output from each node
		.. you'll obviously need to modify this to fit your
		.. needs but this is the general idea
	</xsl:for-each>
</xsl:template>
> 3. Given the hierarchical output from 2, I would like to extract all
> nodes which are children of any of the ancestors of the node with a
> given MenuId (including that node itself)
> Expressed otherwise: the children, siblings, ancestors, uncles and
> great-uncles, great-great-uncles (etc) of the given node.
> Expressed otherwise: all Menus where MenuParentId IN (MenuIds of a given
> node and all of its ancestors)

<xsl:template match="/Menus">
	<xsl:apply-templates select="Menu"/>
</xsl:template>

<xsl:template match="Menu">
<xsl:variable name="MenuId" select="@MenuId"/>
	<xsl:apply-templates select="ancestor-or-self::Menu[@MenuId = $MenuId]" 
mode="MenuId">
		<xsl:with-param name="MenuId" select="$MenuId" />
	</xsl:apply-templates>
</xsl:template>

<xsl:template match="Menu" mode="MenuId">
<xsl:param name="MenuId"/>
... output what you want here ...
<xsl:apply-templates select="*[@MenuId = $MenuId]" mode="MenuId">
	<xsl:with-param name="MenuId" select="$MenuId"/>
</xsl:apply-templates>
</xsl:template>

Untested, but this is the general idea...

Im in the middle of some other code but as soon as I have a chance I will test 
this and see if there are any blaring mistakes and respond back if yes...

Best of luck!

<M:D/>
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> Many thanks for any answers received
>
> Ben Simkins