- 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