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RE: [xsl] replace

2007-06-12 00:54:00
I suspect you want select="'a'" and select="'b'".

Why are you still using XSLT 1.0, as a matter of interest? I can understand
it if you're in a constrained environment, e.g. in a browser, but if you're
using Saxon then that's clearly not the case. Doing string manipulation in
1.0 when you could be using 2.0 seems like pure masochism to me.

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


-----Original Message-----
From: Charles Ohana [mailto:charles(_dot_)ohana(_at_)buongiorno(_dot_)com] 
Sent: 12 June 2007 03:28
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] replace

Hello,
I'm sorry, I know this question has been asked several times 
but ... After searching everywhere I could not get the 
replace function working.
See my code below you'll see what I'm trying to accomplish. 
All I need is a simple working example that is replacing a 
pattern withing a value.
Any hint would be appreciated. I'm using SAXON 6.5.5 (java 
-jar saxon.jar)

Thank you

<?xml version="1.0" encoding="utf-8"?>
<stylesheet xmlns="http://www.w3.org/1999/XSL/Transform";
            xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
            xmlns:str="http://exslt.org/Strings"; version="1.0"
            extension-element-prefixes="str">

<import href="functions/replace/str.replace.xsl"/>
<xsl:template match="/">
<xsl:call-template name="str:replace">
   <xsl:with-param name="string" select="/root/@description" />
   <xsl:with-param name="search" select="a" />
   <xsl:with-param name="replace" select="b" /> 
</xsl:call-template> </xsl:template>

</xsl:stylesheet> 


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