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(_at_)nag(_dot_)co(_dot_)uk>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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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