[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: How to get output XML same as input XML?
Hi Josh,
Try this: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <xsl:copy-of select="/"> </xsl:template> </xsl:stylesheet> Does it give you what you want? If not, please explain how its result varies from your requirement to "[take] the input XML and [turn] exactly what was input into output XML". Depending on exactly what you mean by this, it may or may not be possible to do in XSLT. Examining the output of the above tranformation will tell. There are other stylesheets that will simply copy input to output, but this is the simplest. Cheers, Wendell At 03:49 PM 7/9/2003, you wrote: Hi all - ====================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx Mulberry Technologies, Inc. http://www.mulberrytech.com 17 West Jefferson Street Direct Phone: 301/315-9635 Suite 207 Phone: 301/315-9631 Rockville, MD 20850 Fax: 301/315-8285 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ====================================================================== XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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