Subject: [xsl] Document reordering after applying path expression From: "Max Toro maxtoroq@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Tue, 30 Jun 2015 19:19:58 -0000 |
The following stylesheet: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="2.0" exclude-result-prefixes="#all" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xsl:output method="xml" indent="yes"/> <xsl:template name="main"> <xsl:variable name="a"> <a/> </xsl:variable> <xsl:variable name="b"> <b/> </xsl:variable> <for-each1> <xsl:for-each select="reverse(($a, $b))"> <xsl:copy-of select="."/> </xsl:for-each> </for-each1> <for-each2> <xsl:for-each select="reverse(($a, $b))/*"> <xsl:copy-of select="."/> </xsl:for-each> </for-each2> </xsl:template> </xsl:stylesheet> Outputs: <?xml version="1.0" encoding="UTF-8"?> <for-each1> <b/> <a/> </for-each1> <for-each2> <a/> <b/> </for-each2> Why is the sequence reordered after applying a path expression? Is this conformant with the standard? If yes, what section talks about it? I'm using Saxon 9.6.0.6. Original post: http://stackoverflow.com/questions/31126950 -- Max Toro
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