RE: [xsl] xalan and replace "regex"

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RE: [xsl] xalan and replace "regex"

Subject: RE: [xsl] xalan and replace "regex"
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Thu, 27 Dec 2007 10:54:46 -0000

> Is there any way I can use replace with xalan. 

The replace() function is new in XSLT 2.0, and Xalan only supports XSLT 1.0.
Switch to Saxon.

Michael Kay
http://www.saxonica.com/

Current Thread
[xsl] xalan and replace "regex" Mansour - Thu, 27 Dec 2007 05:02:40 -0400 Colin Paul Adams - Thu, 27 Dec 2007 09:10:37 +0000 Michael Kay - Thu, 27 Dec 2007 10:54:46 -0000 <= Mansour - Thu, 27 Dec 2007 10:09:48 -0400 Abel Braaksma - Thu, 27 Dec 2007 15:15:11 +0100 Mansour - Thu, 27 Dec 2007 10:20:25 -0400 Abel Braaksma - Thu, 27 Dec 2007 15:29:22 +0100

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