Yes, this is exactly what I meant by swithcing the context to document 2,
but I think it can not really be suggested, because you can not switch back
to the cat without rewriting the code. Furthermore <xsl:sort
select="*[name() = $sortparam]"/> is not possible.
Regards,
Joerg
Dimitre Novatchev wrote:
--- "Carter, Will" <WCarter at envestnetpmc dot com> wrote:
Hi,
I have a sorting problem when I use xsl to combine data from 2 xml
files into one html output.
here is my xml file 1 (xml1.xml):
-----------------------
<people>
<person name="george">
<cat>cat-zoro</cat>
<dog>dog-butch</dog>
<fish>fish-jaws</fish>
</person>
<person name="jennifer">
<cat>cat-felix</cat>
<dog>dog-fido</dog>
<fish>fish-moby</fish>
</person>
<person name="simon">
<cat>cat-tom</cat>
<dog>dog-scooby</dog>
<fish>fish-conroy</fish>
</person>
</people>
-----------------------
here is my xml file 2 (xml2.xml):
-----------------------
<people>
<person name="george">
<turtle>turtle-greeny</turtle>
</person>
<person name="jennifer">
<turtle>turtle-browny</turtle>
</person>
<person name="simon">
<turtle>turtle-red</turtle>
</person>
</people>
-----------------------
here is my stylesheet:
-----------------------
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";;
xmlns="http://www.w3.org/tr/REC-html40";; version="2.0">
<xsl:key name="turtleByOwner" match="//turtle" use="../@name" />
<xsl:template match="/">
<table border="1" cellspacing="0" cellpadding="0">
<xsl:for-each select="people/person">
<xsl:sort select="cat"/>
<tr>
<td colspan="7">Person: <xsl:value-of select="@name"/></td>
</tr>
<tr>
<td><xsl:value-of select="cat"/></td>
<td><xsl:value-of select="dog"/></td>
<td><xsl:value-of select="fish"/></td>
<xsl:variable name="ownerName" select="@name"/>
<td>
<xsl:for-each select="document('xml2.xml')">
<xsl:value-of select="key('turtleByOwner', $ownerName)"
</xsl:for-each>
</td>
</tr>
</xsl:for-each>
</table>
</xsl:template>
</xsl:stylesheet>
-----------------------
this correctly produces (sorted by cat):
-----------------------
Person: jennifer
cat-felix dog-fido fish-moby turtle-browny
Person: simon
cat-tom dog-scooby fish-conroy turtle-red
Person: george
cat-zoro dog-butch fish-jaws turtle-greeny
-----------------------
but I want to sort by turtle (I want this output):
-----------------------
Person: jennifer
cat-felix dog-fido fish-moby turtle-browny
Person: george
cat-zoro dog-butch fish-jaws turtle-greeny
Person: simon
cat-tom dog-scooby fish-conroy turtle-red
-----------------------
I changed the sort line to be:
<xsl:sort select="turtle"/>
but it doesn't work, my output is:
-----------------------
Person: george
cat-zoro dog-butch fish-jaws turtle-greeny
Person: jennifer
cat-felix dog-fido fish-moby turtle-browny
Person: simon
cat-tom dog-scooby fish-conroy turtle-red
-----------------------
how can I sort on the turtle column from the document reference?
thanks for any ideas.
will
Hi Will,
Here's a solution, which even does *not* need to produce an
intermediary output in an RTF:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/">
<xsl:variable name="firstRoot" select="."/>
<xsl:for-each select="document('person2.xml')/people/person">
<xsl:sort select="turtle"/>
person: <xsl:value-of select="@name"/>
<xsl:text>
</xsl:text>
<xsl:for-each select="$firstRoot/people/person[(_at_)name =
current()/@name]/*">
<xsl:value-of select="concat(., ' ')"/>
</xsl:for-each>
<xsl:for-each select="*">
<xsl:value-of select="concat(., ' ')"/>
</xsl:for-each>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
When applied to your source xml document, the result is:
person: jennifer
cat-felix dog-fido fish-moby turtle-browny
person: george
cat-zoro dog-butch fish-jaws turtle-greeny
person: simon
cat-tom dog-scooby fish-conroy turtle-red
=====
Cheers,
Dimitre Novatchev.
--
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