Re: [xsl] Problem with Sorting Elements When using "document()" to bring in multiple files

Subject: Re: [xsl] Problem with Sorting Elements When using "document()" to bring in multiple files
From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>
Date: Wed, 03 Dec 2003 10:25:44 -0500
At 2003-12-03 14:46 +0000, Dominic Lisi wrote:
i have this problem, when i try to sort elements which are comming from external files.

Sorting works on the current node list, so you must divine an expression in which all of the nodes to be sorted are selected *at one time* into the current node list.


i have a masterfile.xml which contains multiple tags like the following:

<taxpayer file="A004.xml" id="A004"/>

the file reference links to an XML file which contains all the individuals data for example:

<taxpayer id="A004">
   <name>
       <family>Random_1</family>
       <given>Made_Uo_Dude</given>
   </name>
</taxpayer>

Not a problem with the above, except that your stylesheet approach you described did not consider the current node list. Also, you are resolving your relative URI values in the document() function relative to your stylesheet, and I'm guessing you want them resolved relative to your data, so I changed that as well.


Try the following:

   <xsl:template match="/">
      <alpha-list>
         <xsl:apply-templates select="document( list/taxpayer/@file, . )/
                                      taxpayer">
           <xsl:sort select="name"/>
         </xsl:apply-templates/>
      </alpha-list>
   </xsl:template>

The above will push all of your individual taxpayer nodes in sorted order by the text value of the name node (the concatenation of its descendent text nodes ... that's what you implied by your post).

The reason this works is that the document function will return the root nodes of *all* of the documents addressed by *each* of the nodes passed in the first argument, thus supplying you with a single current node list with all the nodes to be sorted.

I hope this helps.

............................... Ken

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