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

2015-03-18 11:44:33
 
 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

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