[xsl] how to work with xml fragment cont'd

[Jakub dot Valenta at Deio dot net]

Thanks Michael,
I tried this but then I need to specify the whole
DTD:

<!DOCTYPE wrapper [
<!ELEMENT wrapper ANY>
<!ELEMENT ....> // all the elements in First.xml
<!ENTITY frag SYSTEM "First.xml">
]>

Is there any way how to avoid the need to specify the whole DTD, I don't
care about the document type, I just want to include it.

br,

Jakub






"Michael Kay" <michael.h.kay@ntlworld.com>@lists.mulberrytech.com on
02/04/2002 11:57:47 AM

Please respond to xsl-list@lists.mulberrytech.com

Sent by:  owner-xsl-list@lists.mulberrytech.com


To:   <xsl-list@lists.mulberrytech.com>
cc:

Subject:  RE: [xsl] how to work with xml fragment

Simplest is to wrap it into a well-formed document:

<?xml ...?>
<!DOCTYPE wrapper [
<!ENTITY frag SYSTEM "fragment.xml">
]>
<wrapper>&frag;</wrapper>

and then supply this wrapper document as input to the XSLT transformation.

Michael Kay
Software AG
home: Michael.H.Kay@ntlworld.com
work: Michael.Kay@softwareag.com

> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of
> Jakub.Valenta@Deio.net
> Sent: 04 February 2002 09:32
> To: XSL-List@lists.mulberrytech.com
> Subject: [xsl] how to work with xml fragment
>
>
> Hi all,
> I have an xml fragment looking like,
> <a>
> </a>
> <a>
> </a>
> <a>
> </a>
>
> and I would like to process it using XSLT, since it is not valid xml
> document (missing root) I don't know how to force the XSLT
> engine to work
> with it.
> Is there any way how to use it either as an input or to get it using
> document() function.
>
> br,
>
> Jakub
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>

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