Subject: [xsl] xsl:sequence produces error I cant understand From: Kenneth Stephen <marvin.the.cynical.robot@xxxxxxxxx> Date: Sun, 10 Oct 2004 14:44:36 -0500 |
Hi, The following code : <?xml version="1.0"?> <xsl:stylesheet version="2.0" xmlns:ns="urn:dummyNamespace" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" /> <ns:root> <ns:child attr="x" /> <ns:child attr="y" /> </ns:root> <xsl:template match="/"> <xsl:for-each select="document('')/xsl:stylesheet/ns:root/ns:child"> <xsl:if test="@attr = 'y'"> <xsl:sequence select="@attr" /> </xsl:if> </xsl:for-each> </xsl:template> </xsl:stylesheet> ....when invoked on itself produces : bash-2.05b$ java net.sf.saxon.Transform seq.xsl seq.xsl Error at if on line 15 of file:/C:/cygwin/home/zaphod/code/seq.xsl: Cannot write an attribute when there is no open start tag Transformation failed: Run-time errors were reported Can someone help me understand my error please? Thanks, Kenneth
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