Subject: [xsl] Bug in Xalan 2.4.1? From: glaprade <glaprade@xxxxxxxxxx> Date: Tue, 21 Jan 2003 12:11:57 -0700 |
I am using the xslt-process module for Emacs and am attempting to produce an example as to why comparing 2 node sets with the = or != operators is problematic. xslt-process allows me to switch between Xalan 2.4.1 and Saxon 6.5.2 for my XSLT processor. I believe I've discovered a bug in Xalan 2.4.1. XML FILE: <Employees> <Employee status="fulltime" pay="salaried"> <FirstName>Jeff</FirstName> <LastName>Smith</LastName> <SSN>323345333</SSN> </Employee> <Employee status="parttime" pay="hourly"> <FirstName>Mike</FirstName> <LastName>Young</LastName> <SSN>222176543</SSN> </Employee> <Employee status="parttime" pay="salaried"> <FirstName>George</FirstName> <LastName>Castanza</LastName> <SSN>121345123</SSN> </Employee> </Employees> XSL FILE: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" indent="yes" encoding="UTF-8"/> <xsl:template match="/"> <root> <xsl:variable name="NodeSet1" select="/Employees/Employee[1]"/> <xsl:variable name="NodeSet2" select="//Employee"/> <xsl:if test="$NodeSet1 = $NodeSet2"> <Equal/> </xsl:if> <xsl:if test="$NodeSet1 != $NodeSet2"> <NotEqual/> </xsl:if> </root> </xsl:template> </xsl:stylesheet> With Saxon 6.5.2 selected I get: <?xml version="1.0" encoding="UTF-8"?> <root> <Equal/> <NotEqual/> </root> With Xalan 2.4.1 selected I get: F:/stuff/XSLT/equal_notequal_gotcha.xsl:13:38: Fatal error: Unknown error in XPath which points to this line: <xsl:if test="$NodeSet1 != $NodeSet2"> MSXML 4 will output the same as Saxon. I believe this is a bug, but wanted to ensure that there is nothing wrong with my not equal comparison. Thanks, Greg XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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