Subject: Re: [xsl] EXSLT replace to XSLT 2.0 replace From: "Imsieke, Gerrit, le-tex" <gerrit.imsieke@xxxxxxxxx> Date: Wed, 14 Apr 2010 10:56:36 +0200 |
So either: <xsl:value-of select="replace( .,
'(?:(.+@.+\.[a-zA-Z]+)|([\p{Lu}|\p{Ll}]\p{Ll}+)-([\p{Ll}]+))', '$1$2$3', '!' )"/> or: <xsl:value-of select="replace( .,
'((.+@.+\.[a-zA-Z]+)|([\p{Lu}|\p{Ll}]\p{Ll}+)-([\p{Ll}]+))', '$2$3$4' )"/>
Omit the 'g', which replace() does anyway.
Omit the '(?: ... )" from the regular expression. (This is uncounted grouping, which isn't important here.
The | in (\p{Lu}|\p{Ll}] looks suspicious. These are alternatives, anyway, within []
-W
On Wed, Apr 14, 2010 at 9:48 AM, Szabo, Patrick (LNG-VIE) <patrick.szabo@xxxxxxxxxxxxx> wrote:Hi,
I'm using Saxon 9 and XSLT 2.0
I'm currently updateing a few 1.0 Stylesheets to 2.0. In one of those Stylesheets i've found this:
<xsl:value-of select="regexp:replace(., '(?:(.+@.+\.[a-zA-Z]+)|([\p{Lu}|\p{Ll}]\p{Ll}+)-([\p{Ll}]+))', 'g', '$1$2$3')"/>
Is there a way to to that in 2.0 without EXSLT ?! I - of course - wanted to try it with replace() but i cant seem to figure it out.
best regards...
Patrick Szabo XSLT-Entwickler
Tel.: +43 (1) 534 52 - 1573 Fax: +43 (1) 534 52 - 146
Patrick.Szabo@xxxxxxxxxxxxx
LexisNexis Verlag ARD Orac GmbH& Co KG Marxergasse 25, 1030 Wien FN 8333f, Handelsgericht Wien http://www.lexisnexis.at/
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