Re: [xsl] sorting information from multiple files

Subject: Re: [xsl] sorting information from multiple files
From: "james walker" <jameswalkerandy@xxxxxxxxxxx>
Date: Thu, 15 Jan 2004 15:57:45 +0000
Hello,
Its a little more complicated now as i managed to signififcantly reduce the amount of code by using nsindex.
<xsl:for-each select="child::category">
<xsl:variable name="catnamemod" select="translate(name,' ','_')" />
<xsl:for-each select="document($nsindex/*[name()=$catnamemod])/root/child::*">
<xsl:if test="@topfive=true()">
<xsl:variable name="url" select="url" />
<li><a href="{$url}"><xsl:value-of select="title" /></a></li>
</xsl:if>
</xsl:for-each>
</xsl:for-each>
Basically in the second for-each there could be anything from 1-9 documents which are only known during the translation. The content form these files needs to be alphabetically ordered and i wasnt sure how to do it?
nsindex refers to a node set of urls where the name of teh variables are equal to the different catnamemods...



From: David Carlisle <davidc@xxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] sorting information from multiple files
Date: Thu, 15 Jan 2004 15:38:53 GMT

> i have 4 for-each loops which gather specific info from each file:

why not just have one:

<xsl:for-each select="document(http:..doc1..)/foo/bar |
                      document(http:..doc2..)/x/y |
                      document(http:..doc3..)/a/b |
                      document(http:..doc4..)/zz/c">
<xsl:sort select="....



--
http://www.dcarlisle.demon.co.uk/matthew

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