Subject: [xsl] xsl:analyze-string and line break From: "Mathieu Malaterre" <mathieu.malaterre@xxxxxxxxx> Date: Wed, 10 Oct 2007 12:28:55 +0200 |
Hello, I am trying to do a regex on an expression with line breaks, for some reason '.' does not include line break. I also tried [.\n]* to say anything including line break, with no luck. xml file is: <?xml version="1.0"?> <description>Sex of the named patient. Enumerated Values: M = male F = female O = other</description> and xsl file is: <?xml version="1.0"?> <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> <xsl:output method="xml" indent="yes"/> <xsl:template name="parse-enum"> <xsl:param name="text"/> <xsl:analyze-string select="$text" regex="\n"> <xsl:matching-substring> <!--br/--> </xsl:matching-substring> <xsl:non-matching-substring> <enum> <xsl:value-of select="."/> </enum> </xsl:non-matching-substring> </xsl:analyze-string> </xsl:template> <xsl:template match="/description"> <xsl:analyze-string select="." regex=".*Enumerated Values:([.\n]*)"> <xsl:matching-substring> <xsl:value-of select="regex-group(1)"/> </xsl:matching-substring> </xsl:analyze-string> </xsl:template> </xsl:transform> Thanks, -- Mathieu
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