RE: sort with different files

You are only retrieving the last name from one file. Your XPath: 
document(./url/@path)/lastname\
seems to be coming from the context of pers_data, in which case there will only be one url loaded.

You could try something like this:

<xsl:template match="/root">
  <xsl:apply-templates/>
</xsl:template>

<xsl:template match="node">
  <!-- Load all the pers_data urls into a variable for access --> 
  <xsl:variable name="data" select="document(pers_data/url/@path)"/>
  
  <xsl:for-each select="pers_data[@status='active']">
  <xsl:sort select="$data/pers_data[@id = current()/@id]/lastname"/>
    <lastname>
    <xsl:value-of select="$data/pers_data[@id = current()/@id]/lastname"/>
    </lastname>
  </xsl:for-each>
</xsl:template>

Josh

-----Original Message-----
From: Markus Hanel [mailto:markus.hanel@xxxxxx]
Sent: Thursday, April 22, 2004 4:27 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  sort with different files


I have tried it with the document, but the problem is that i work with the
file students.xml where the pers_data nodes included. But the pers_datas are
destaged with a url. Which document must I select by the xsl element sort?

> you can do a sort using the document function if thats what you mean
> 
> -----Original Message-----
> From: Markus Hanel [mailto:markus.hanel@xxxxxx]
> Sent: 22 April 2004 12:16
> To: xsl mailinglist
> Subject:  sort with different files
> 
> 
> 
> Hello,
> i have a table, where I want to list some students given by there
> pers_data
> in the file students.xml. Therefore I made a file a_format.xml to generate
> columns automaticaly which are given as child of the node interviewer. In
> the file students.xml the pers_data is destaged with a url.
> My problem is now to sort the students by their lastname. The difficult is
> that there are different files!
> I hope someone can help me.
> Many thanks
> markus
> 
> file a_format.xml
> <format>
> ...
>   <interviewer>
>     <show>id</show>
>     <show>lastname</show>
>     <show>surname</show>
>     <show>userid</show>
>     <show>sex</show>
>     <show>birth</show>
>     <show>class</show>
>   </interviewer>
> ...
> </format>
> 
> file students.xml
> <root>
> ...
> <node type="interviewee" status="active">
>   <pers_data status="active" task="interviewee" id="3">
>     <url path="/qpers_data/3.xml" proto="file">
>     </url>
>   </pers_data>
>   <pers_data status="active" task="interviewee" id="4">
>     <url path="/qpers_data/4.xml" proto="file">
>     </url>
>   </pers_data>
> </node>
> ...
> </root>
> 
> file 3.xml
> 
> <pers_data task="interviewee" id="3" status="active">
>   <proto>file</proto>
>   <type>interviewee</type>
>   <surname>name</surname>
>   <lastname>name</lastname>
> ...
> </pers_data>
> 
> xsl stylesheet:
> 
> <table>
>   <tr>
> <!--generate columns automaticaly which are given in a_format.xml -->
>     <xsl:for-each
> select="document('/qxml/a_format.xml')/format/interviewer/child::*">
>       <xsl:variable name="show_node" select="." />
>       <th><xsl:value-of
>
select="../../style_body/style_display/translation/*[name()=concat('trans_',
> $show_node)]"
> /></th>
>     </xsl:for-each>
>   </tr>
> <!-- list the pers_data of the file students.xml -->
>   <xsl:for-each select="$self_node/pers_data[attribute::status='active']">
> <!-- sort ???????? -->
>   <xsl:sort select="document(./url/@path)/lastname" />
>   <xsl:variable name="pers_data_file" select="document(./url/@path)" />
>   <xsl:variable name="pos" select="position()" />
>       <tr>
>         <xsl:for-each
> select="document('/qxml/a_format.xml')/format/interviewer/child::*">
>         <xsl:variable name="show" select="." />
>           <td>
>             <xsl:value-of
> select="$self_node/pers_data[$pos]/*[name()=$show]
> | $pers_data_file/pers_data/*[name()=$show]" />
>           </td>
>         </xsl:for-each>
>       </tr>
>   </xsl:for-each>
> </table>