RE: How to find all parents having the same child?

Jarno,

that was exactly what I much needed. 

Thanks a lot.
bjoon

 
--- Jarno.Elovirta@xxxxxxxxx wrote:
> Hi,
> 
> > Here an example:
> > 
> > <Quries>
> >  <Query name="query1">
> >    <Query name="query2">
> >      <Table name="table1"/>
> >      <Table name="table2"/>
> >      <Query name="query3">
> >        <Table name="table3"/>
> >      </Query>
> >    </Query> 
> >  </Query>
> >  <Query name="query4">
> >    <Table name="table1"/>
> >    <Table name="table2"/>
> >    <Table name="table3"/>
> >  </Query> 
> > </Queries>
> > 
> > Given that, I would like to extract all parents
> for
> > each leaf like the following:
> > 
> > table3: query4, query3
> > table1: query4, query1
> 
> did you mean
> 
>   table1: query4, query2
> 
> as query1 is not a direct parent of table1.
> 
> > query3: query2
> > query2: query1
> > 
> > An example from you on how to get that
> cross-reference
> > list would help me a lot.
> 
> Hopefully this will get you started.
> 
>   <xsl:key name="x" match="Table | Query"
> use="@name"/>
>   <xsl:output method="text"/>
>   <xsl:template match="Queries">
>     <xsl:for-each
> select="descendant::*[generate-id(.) =
> generate-id(key('x', @name))]">
>       <xsl:if test="key('x', @name)[parent::Query]">
>         <xsl:value-of select="@name"/>: <xsl:text/>
>         <xsl:for-each select="key('x', @name)">
>           <xsl:if test="not(position() = 1)">,
> </xsl:if>
>           <xsl:value-of select="../@name"/>
>         </xsl:for-each>
>         <xsl:text>
</xsl:text>
>       </xsl:if>
>     </xsl:for-each>
>   </xsl:template>
> 
> Cheers,
> 
> Jarno - Chris C: Vengeance Is Mine
> 
>  XSL-List info and archive: 
> http://www.mulberrytech.com/xsl/xsl-list
> 


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