RE: RE: XPath expression confusion

Thanks.
--
Charles Knell
cknell@xxxxxxxxxx - email

-----Original Message-----
From:     Xia Li <xli@xxxxxxxxxxxxx>
Sent:     Thu, 12 Jan 2006 15:39:58 -0800
To:       <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Subject:  RE:  XPath expression confusion

<xsl:when test="not(/system/components/component/requirement/@number .)">
	<requirement><xsl:value-of select="." /></requirement>
</xsl:when?

That means if there exists an item in the sequence selected by
"/system/components/component/requirement/@number" that is not equal to
the current value of <number> in the iteration, the condition is always
true. That is why you output all the value of <number> elements.

File A
-------
<requirements>
 <requirement>
   <number />
     <requirement>
       <number />
       And so on to an arbitrary depth
     </requirement>
 </requirement>
</requirements>

File B
-------
<system>
 <components>
   <component>
     <requirement number="" />
     <requirement number="" />
     <requirement number="" />
   </component>
 </components>
</system>

There will be an arbitrary number of components, each with an arbitrary
number of requirements.

My goal is to produce a document containing all the requirement/number
elements from File A that don't have a matching requirement/@number in
File B.

<xsl:variable name="all-reqs"
select="document('requirements.xml')/requirements//requirement/number"
/>

 <xsl:template match="/">
   <un-assigned-reqs>
     <xsl:for-each select="$all-reqs">
       <xsl:choose>
         <xsl:when
test="/system/components/component/requirement/@number = ." />
         <xsl:otherwise>
           <requirement><xsl:value-of select="." /></requirement>
         </xsl:otherwise>
       </xsl:choose>
     </xsl:for-each>
   </un-assigned-reqs>
 </xsl:template>

It was my belief that the XPath expression in the test would search all
the number attribute values in the document for a value that matched the
value of the current iteration of the loop.
Instead, it seems that there are no matches, as I get one element output
for each iteration of the loop.

--
Charles Knell
cknell@xxxxxxxxxx - email