Re: XPath expression problem

[Joerg Heinicke]

Hello Jacques,

this is a FAQ about XSLT. You find the solution nearly everywhere (and 
below), e.g. at http://www.jenitennison.com/xslt/grouping/muenchian.xml. 
If not, there are specific XSLT list, the best probably at 
http://www.mulberrytech.com/xsl/xsl-list/

Enough admonished :)

Now the answer ...

On 16.04.2004 17:32, [EMAIL PROTECTED] wrote:
Hello,

I have a problem with an XPath. My XML-Document that i want to format
with my XSLT-stylesheet is a storylist which contains a number of
stories which contains a title and a body-element. The body element
contains an attribute called language. Here is a excerpt of my DTD:
<!ELEMENT storylist  (story*) >
<!ELEMENT story  (body, title) >
<!ATTLIST story
            	   id  ID    #REQUIRED 
            	   xreftext  CDATA    #IMPLIED  
            	   author CDATA #IMPLIED 
            	   category CDATA #IMPLIED>
<!ELEMENT body  (#PCDATA | xref | seealso | url)* >
<!ATTLIST body
                language (en|de|es|it|jp)        "de"> 
...

I'm using a key 'language-index' to get all messages of the same
languages:
  <xsl:key name="language-index" match="body" use="@language"/>
I want to iterate through those stories after i sort the by the
author-attribute that is in the body of the story i do this inside
<xsl:template match="/">:        
<xsl:for-each select="key('language-index', $targetLanguage)">
        	<xsl:sort select="ancestor::story/@author"/>

Now i want to print some text only if the author of the story has
changed from the preceding story. 
Stating your problem is of course important for finding solutions. With 
this the Mulberry XSL list mentioned above can help you a lot as there 
are many mind-readers subscribed ;-) What you want to do is grouping the 
stories by authors. The Muenchian Grouping is the common approach:

<xsl:key name="author-index" match="story" use="@author"/>

<xsl:template match="/storylist">
  <table>
    <xsl:apply-templates select="story[generate-id() = 
generate-id(key('author-index', @author))]" mode="group">
      <xsl:sort select="@author"/>
    </xsl:apply-templates>
  </table>
</xsl:template>

<xsl:template match="story" mode="group">
  <!-- first story of this author => group starter -->
  <tr>
    <th><xsl:value-of select="@author"/></th>
  </tr>
  <!-- select all stories of this author -->
  <xsl:apply-templates select="key('author-index', @author)"/>
</xsl:template>
<xsl:template match="story">
  <tr><td><xsl:value-of select="title"/></td></tr>
</xsl:template>
This misses now the handling of language, but I think you can add it 
easily. From you original code I assume you only need onle language at a 
time, so you should add this to the key. If you need instead an 
additional grouping by language it gets a bit more complicated. You have 
to setup a second key concatenated from the author and the language.

My problem is that i don't know how to access the author-attribute of
the preceding sibling. I have tried a couple of different syntaxes but
none of them worked what i want is someting like this:
   preceding-sibling::story[1]/@author
but that returns NULL.
The problem is, this can not work. Even if you sort the nodes the order 
of them when using axes like preceding-sibling is the original one.

i can access the author associate with the current body-node like this:
   ancestor::story[1]/@author
You should not use ancestor, but parent axis or its short form: 
parent::story/@author or ../@author

Joerg

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