Subject: RE: [xsl] xsl:analyze-string and line break From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Wed, 10 Oct 2007 14:12:21 +0100 |
I've no idea of the history that caused "." to have this meaning, but your observation is correct. To make "." match a newline, use flags="s" on the xsl:analyze-string instruction. Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Mathieu Malaterre [mailto:mathieu.malaterre@xxxxxxxxx] > Sent: 10 October 2007 11:29 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] xsl:analyze-string and line break > > 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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