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[(_at_)status='active']">
<xsl:sort select="$data/pers_data[(_at_)id = current()/@id]/lastname"/>
<lastname>
<xsl:value-of select="$data/pers_data[(_at_)id = current()/@id]/lastname"/>
</lastname>
</xsl:for-each>
</xsl:template>
Josh
-----Original Message-----
From: Markus Hanel [mailto:markus(_dot_)hanel(_at_)gmx(_dot_)at]
Sent: Thursday, April 22, 2004 4:27 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] 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(_dot_)hanel(_at_)gmx(_dot_)at]
Sent: 22 April 2004 12:16
To: xsl mailinglist
Subject: [xsl] 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>
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