Subject: [xsl] How to get output XML same as input XML? From: Joseph Kesselman <keshlam@xxxxxxxxxx> Date: Wed, 9 Jul 2003 17:13:29 -0400 |
Canonical Identity Transformation, right out of the XSLT spec: <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:stylesheet> This does not guarantee that the output document exactly matches the input; details not considered meaningful to XML or to XSLT's data model (eg, the order of attributes) may be changed. But that falls in the category of "a difference which makes no difference is no difference" -- if you don't consider those alterations acceptable, don't put the document through generic XML-based processing. ______________________________________ Joe Kesselman, IBM Next-Generation Web Technologies: XML, XSL and more. "The world changed profoundly and unpredictably the day Tim Berners Lee got bitten by a radioactive spider." -- Rafe Culpin, in r.m.filk XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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