It is my understanding that Java' regular expression builtin emulates
'pcre' pretty closely.
To escape spacial characters that have special meaning in a regular
expression, defining it as a character class (using the square bracket
notation) generally works
ie. if you want to match a question mark at the beginning of a line,
use: "^[?].*$"
On Wed, Jul 23, 2014 at 3:55 PM, mike(_at_)saxonica(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Exclamation mark is not a special character in XPath regular expressions,
and there does not need to be (and must not be) escaped.
Negative lookaheads are not supported in the XPath regular expression
dialect.
You can't assume that all regular expression dialects are the same.
Michael Kay
Saxonica
Dear All,
I am using xsl:analyze-string to retrieve and replace punctuation,
however, I got the following error:
Error in regular expression: net.sf.saxon.trans.XPathException: Syntax
error at char 6 in regular expression: Escape character '!' not allowed.
How should I escape and match '?' and '!' ? I am also using a negative
look-ahead, why isn't that working?
Here is a sample from my code, thanks,
Gabor
<xsl:template match="//TEI:p//text()[ not
((parent::TEI:note)|(parent::TEI:hi)|(parent::TEI:date))]">
<xsl:analyze-string select="." regex="(\.|\!|\?)(?!\)|\.|\d|\w)">
<xsl:matching-substring>
<xsl:element name="seg"
namespace="http://www.tei-c.org/ns/1.0";><xsl:value-of
select="."/></xsl:element>
</xsl:matching-substring>
<xsl:non-matching-substring>
<xsl:value-of select="."/>
</xsl:non-matching-substring>
</xsl:analyze-string>
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