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further question regarding the use of Muenchian Method
- From: "=?big5?B?qkwgpGyq5A==?=" <minikittygo at hotmail dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Mon, 15 Apr 2002 17:43:18 +0000
- Subject: [xsl] further question regarding the use of Muenchian Method
- Reply-to: xsl-list at lists dot mulberrytech dot com
hi
i was wondering whether it is possible to group the content given that they
belongs to a separate structure (i.e.
<booklist>
<adventure>
<author>author1</author>
<title>1</title>
<adventure>
<adventure>
<author>author2</author>
<title>2</title>
<adventure>
<adventure>
<author>author1</author>
<title>3</title>
<adventure>
<sci-fi>
<star-trek>
<author>author1</author>
<title>6</title>
</star-trek>
<star-trek>
<author>author4</author>
<title>9</title>
</star-trek>
</sci-fi>
</booklist>
transform to
<booklist>
<books>
<author>author1</author>
<title>1</title>
<title>3</title>
<title>6</title>
</books>
<books>
<author>author2</author>
</title>2</title>
</books>
...etc
since there are two different 'forms' of author 'tags' (i am sure there is
a better description) (i.e adventure/author and sci-fi/star-trek) i could
not generate the appropiate key for mucheun method (well at least i am not
certain how) but is there are a way to do so?
going deeper into the question, is it possible to do group the content even
the content contain in a separate file (i.e. 1.xml and 2.xml)? do i simply
ise xsl:include?
Thanks for your patiences
Many thanks
Regards
Kit
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