Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?

Subject: Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?
From: "Michael PG" <xrow@xxxxxxx>
Date: Tue, 05 Oct 2004 16:35:53 +0000

no output at all. only <Documents></Documents> after adding

   <xsl:template match="Documents">
	<xsl:apply-templates select="*[@filter='' or @filter=$filter]"/>

input XML that I try to parse looks like:

	<Document chapter="1" title="title 1" href="file1.xml" filter="">
		<Article title="1.1" info="sub" filter="food"/>
		<Article title="1.2" info="main" filter="food"/>
	<Document chapter="2" title="title 2" href="file2.xml" filter="drink">
		<Article title="2.1" info="sub" filter="drink"/>
		<Article title="2.2" info="main" filter="food"/>


From: David Carlisle <davidc@xxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Re: [xslt transform & grouping] Using the Muenchian Method?
Date: Tue, 5 Oct 2004 17:23:26 +0100

That is one of the issues.

Is there a way to solve this ?

isn't that where we came in?

 <xsl:template match="Documents">
	<xsl:apply-templates select="*[@filter=$filter]"/>

says apply to children of Documents that have a filter attribute equal to
$filter. I think what you are saying is that you want to apply templates
to Document elements that have filter attribute equal to "" or no filter
attribute or have attribute equal to $filter.

It's easier to say that in XPath than in English

<xsl:template match="Documents"> <xsl:apply-templates select="*[@filter='' or @filter=$filter]"/>

(@filter='' will be true if it is there and empty or if it is not there
at all)


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