- From: <bugzilla@wiggum.w3.org>
- Date: Mon, 17 Aug 2009 19:14:13 +0000
- To: public-qt-comments@w3.org
http://www.w3.org/Bugs/Public/show_bug.cgi?id=7354 Summary: Meaning of "part of a sequence constructor" Product: XPath / XQuery / XSLT Version: Recommendation Platform: PC URL: http://www.w3.org/TR/xslt20/#forwards OS/Version: Windows XP Status: NEW Severity: trivial Priority: P5 Component: XSLT 2.0 AssignedTo: mike@saxonica.com ReportedBy: zongaro@ca.ibm.com QAContact: public-qt-comments@w3.org In section 3.9 of XSLT 2.0,[1] the third bullet of the second list begins, "if an element in the XSLT namespace appears as part of a sequence constructor, and XSLT 2.0 does not allow such elements to appear as part of a sequence constructor," and then goes on to describe the behaviour, depending on whether the element has any xsl:fallback children. Now consider the following fragment: <xsl:template match="/"> <xsl:choose version="8.5"> <xsl:when test="condition"> Do something </xsl:when> <xsl:otherwise> Do something else </xsl:otherwise> <xsl:finally> <xsl:fallback/> </xsl:finally> </xsl:choose> </xsl:template> Presumably the xsl:finally here is not considered to be "part of the sequence constructor," despite the fact that it is part of the content of a node that is part of the sequence constructor. If that interpretation is correct, I would suggest adding the following sentence after the second list to decrease the possibility of confusion: "A node is considered to be part of the sequence constructor if it is one of the sequence of sibling nodes of which the sequence constructor is composed. Descendants of such a node are not part of the sequence constructor." [1] http://www.w3.org/TR/xslt20/#forwards -- Configure bugmail: http://www.w3.org/Bugs/Public/userprefs.cgi?tab=email ------- You are receiving this mail because: ------- You are the QA contact for the bug.
Received on Monday, 17 August 2009 19:14:22 UTC