Subject: Re: [xsl] null stylesheet? From: "Kulas, Jack" <Jkulas@xxxxxxxxxx> Date: Thu, 1 May 2014 03:10:03 -0000 |
A most timely thread for me. I just encountered the "simplified stylesheet." I ran into it in the first design-pattern example of Chapter 17, "Stylesheet Design Patterns," in Michael's "XSLT2.0 and XP2ath 2.0, 4th ed." (BTW, when can we expect the XSLT 3.0 book?] The stylesheet there starts with <html xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xsl:version="2.0"> and has no xsl:stylsheet element. It was strange to see instances of XSLT elements xsl:for-each and xsl: value-of in that stylesheet without the xs:stylesheet element as the outermost element. I since learned that as long as the XSLT namespace is declared and there is an xsl:version attribute in its outermost element, which can't be in the XSLT namespace, it's an XSLT stylesheet! Thus the following also counts as an XSLT stylesheet: <html xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xsl:version="2.0"/> --Jack Kulas LSI, Inc Jacksonvlle, FL -----Original Message----- From: Michael Kay [mailto:mike@xxxxxxxxxxxx] > > and that works perfectly well. But then I tried the slightly simpler > > <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > version="1.0"/> Yes, that's a legal stylesheet. > > But Saxon-HE 9.5.1.3J run from the commandline complains that "The > supplied file does not appear to be a stylesheet" > (this is an error, not a warning; # XTSE0150). > I suspect you got the command line arguments wrong. This message from Saxon means that the file supplied as the stylesheet document did not have xsl:stylesheet or xsl:transform as its root. In fact more specifically I think it means the local name of the root element was not "stylesheet" or "transform" - you get a different message if only the namespace is wrong. Michael Kay Saxonica
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