Maybe because text files can be UNIX escaped (\n), dos (\r\n) or mac
(can't remember).
Anyway regex are awsome !
2 cents,
-Mathieu
On 10/10/07, Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:
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(_dot_)malaterre(_at_)gmail(_dot_)com]
Sent: 10 October 2007 11:29
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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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