worked great; thx so much!!

-- 

Thanks,
David

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
David R. Schwartz, Ph.D.
Bellevue, WA
drs58@m...
"Oleg Tkachenko [MVP]" <oleg@NO!SPAM!PLEASEtkachenko.com> wrote in message
news:uc$m1dcXEHA.1036@T......
> David Schwartz wrote:
>
> > XML:
> > <root>
> > <team_member id="GW">George Washington</team_member>
> > <team_member id="TJ">Thomas Jefferson</team_member>
> > <issue id="i.0034.1">
> >   <description>first issue</description>
> >   <team_member_assoc member="GW"/>
> > </issue>
> > <issue id="i.0034.2">
> >   <description>2ndissue</description>
> >   <team_member_assoc member="GW"/>
> >   <team_member_assoc member="TJ"/>
> > </issue>
> > </root>
> >
> > DESIRED OUTCOME:
> > I want to create a list for each team_member of the issues with which
> > they're associated.
>
> Something like this:
>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
>    <xsl:key name="issuesByMember" match="root/issue"
> use="team_member_assoc/@member"/>
>    <xsl:template match="/">
>      <table>
>        <tr>
>          <th>Team Member</th>
>          <th>Issues</th>
>        </tr>
>        <xsl:apply-templates select="root/team_member"/>
>      </table>
>    </xsl:template>
>    <xsl:template match="team_member">
>      <tr>
>        <td>
>          <xsl:value-of select="."/>
>        </td>
>        <td>
>          <xsl:for-each select="key('issuesByMember', @id)">
>            <xsl:value-of select="description"/>
>            <br/>
>          </xsl:for-each>
>        </td>
>      </tr>
>    </xsl:template>
> </xsl:stylesheet>
>
> -- 
> Oleg Tkachenko [XML MVP]
> http://blog.tkachenko.com