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 16:16:54 +0000
Yeah, it works by adding defining a key:

<xsl:key name="by-info" match="Article" use="@info"/>

<xsl:param name="filter" select="'food'"></xsl:param>


But, I still got the problem with traversing the complete tree for select=food.


E.g. If the parent node element Document does not have filter-attribute defined or if it's equal to zero, that is filter="", than the complete algorithm stops traversing the tree.
That is one of the issues.


Is there a way to solve this ?

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



Thank you !





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 16:06:51 +0100


> Use this XSLT to genterate new XML file by using specified XML input
> file:

clearly that will generate an error as it uses a kety that is not
defined, as I pointed out in the message that you quoted.
As such it will generate no output.

If you fix that then you will get no error

and if you fix     <xsl:param name="filter" select="food"></xsl:param>
to have 'food' then you will get the output that I also posted in that
message, assuming you have a conforming processor.

If you leave $filter with that value you'll just get an empty Documents
element as $filter would be an empty node set as previously explained.

David



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