Subject: Re: [xsl] How to get output XML same as input XML? From: David Carlisle <davidc@xxxxxxxxx> Date: Mon, 14 Jul 2003 13:06:55 +0100 |
If I read your samples correctly, what you are trying to say is that your stylesheet copied an attribute interact:isParameter to the result and did not place a suitable namespace declaration in the linearisation of the result tree. If so that does not show a limitation in XSLT, it is simply a bug in your processor. me>I suspect your terminology here is non standard. a well formed document me>can only have one child element of the root node. > Correct. I am feeding two child elements of my root node separately to > separate callings of the XSL processor. As I suspected, they are not children of the root node, they are children of the topmost element, root nodes are not elements in Xpath. > So the following will be picked up by the >processor as illegal: > ><xsl:template name="openElement"> > <myElement> ></xsl:template> That of course isn't an XSLT restriction, the input would not be passed to XSLT at all as it wouldn't get past teh XML parser. > If my output method is text then this problem does not happen. that input would be an error to any XML application, and in particular to XSLt, whatever output type. > Otherwise > I have to write more code which serializes every element into text, and I > would like to have as much of that done for me as I can. It's only a couple of templates, or a single extension function, both of which can easily be found in the archives, the work is already done for you. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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