Re: [xsl] Muenchian work if more than one value is present

["Liam R. E. Quin liam@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

On Fri, 2015-03-20 at 12:56 +0000, Russell Urquhart russurquhart1@xxxxxxxxxxx wrote:
> As for moving towards XSLT 2.0, i have been wanting to get there but 
> feel i first have to fully understand XSLT 1.0 first! But i'm 
> getting there!

"I can't move to these new-fangled automobiles until I understand how 
to make my own cartwheel with a spokeshave!" :-)

Seriously, I'd say, build some tests and try moving as soon as you can.

Liam


> 
> Thanks to everyone here for your time and patience!
> 
> 
> Russ
> 
> 
> 
> On Thu, Mar 19, 2015 at 07:34:23PM +0000, Wendell Piez 
>  wapiez@xxxxxxxxxxxxxxxxxxxx:
> > Russ,
> > 
> > Yeah -- the for-each pulls all the errata_section elements that 
> > happen to be the first one of the entire set with the same value 
> > for module_impacted -- except that since you've added 
> > module_impact siblings, it will be those that are the first one of 
> > the set of all errata_section with *any* of the same 
> > module_impacted values. The key can work many-to-one, and it does. 
> > This means in principle a module_impacted can be dropped (if it 
> > never appears outside parents that are also bound to other modules 
> > mentioned earlier).
> > 
> > To use Muenchian grouping, write your key to the module_impacted 
> > elements themselves (by their values), not their parents. Then
> > 
> > for-each select="//module_impacted[generate-id() =
> >    generate-id(key('module_impact_mentions',string(.))[1])]">
> > 
> > to iterate over these values (or rather, over a set of node 
> > proxies for them).
> > 
> > Or ditch the Muenchian grouping and join the modern world - XSLT 
> > 2.0.
> > 
> > Cheers, Wendell
> > 
> > 
> > On Wed, Mar 18, 2015 at 12:45 PM,  russurquhart1@xxxxxxxxxxx<
> > xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> > > 
> > >  Hi All,
> > > 
> > > I have a question about using Muenchian Method in a specific 
> > > situation.
> > > 
> > > I have an xml source file like the following:
> > > 
> > > <errata_section id="i875" 
> > > errata_type="bug"><title>Title</title><description> <para> 
> > > Following a warm Reset </para></description><devices_impacted> 
> > > <device_name>VAZER</device_name></devices_impacted><module_impacted>Boot</module_impacted></errata_section>
> > > 
> > > 
> > > The existing xsl, that i am supporting, but didn't write, uses 
> > > the following key:
> > > 
> > > <xsl:key name="module-index" match="errata_section" 
> > > use="module_impacted"/>
> > > 
> > > And they have used the following Muenchian Method algorithm to 
> > > go through all the unique module_impacted elements.
> > > 
> > > 
> > > <xsl:for-each select="//errata_section[generate-id(.)=generate-
> > > id(key('module-index', module_impacted)[1])]"> <xsl:sort 
> > > select="module_impacted"/>
> > > <!-- Determine module_impacted elements, for given device_name, 
> > > and output table. -->
> > > 
> > > </xsl:for-each>
> > > 
> > > Everything went fine until, over time, we have added additional 
> > > module_impacted siblings to the errata_section element like this:
> > > 
> > > <errata_section id="i876" 
> > > errata_type="bug"><title>Title</title><description> <para> 
> > > Following a warm Reset </para></description><devices_impacted> 
> > > <device_name>VAZER</device_name></devices_impacted><module_impacted>Boot</module_impacted><module_impacted>Power-
> > > On</module_impacted><module_impacted>DMA</module_impacted></errata_section>
> > > 
> > > 
> > > Once this happened we started to see that not all 
> > > module_impacted elements were being found for the given 
> > > device_name. My question is, is this happening because the 
> > > element we are doing the generate-id function on, in some cases 
> > > would have multiple siblings in a given errata_section element? 
> > > Most of the examples i have seen of using Muenchian Method are 
> > > keying on a unique elment value that doesn't have any similar 
> > > named siblings.
> > > 
> > > I'm just trying to understand if this structure would work at 
> > > all. Thanks for any info you can provide and thank you Martin 
> > > Honnen for your help so far!
> > > 
> > > Russ