Re: Re: [xslt transform & grouping] Using the Muenchia

> 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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