Re: [xsl] replace

Subject: Re: [xsl] replace
From: "Charles Ohana" <charles.ohana@xxxxxxxxxxxxxx>
Date: Tue, 12 Jun 2007 08:53:06 -0400
I want to replace "a" by "b" inside "/root/@description" . I'm using XSLT 1.0 as I copied this code example from the web. it's not working with 2.0 either anyways. I'm not restricted to any version. All I need is to be able to replace string .
Thank you

----- Original Message ----- From: "Michael Kay" <mike@xxxxxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Tuesday, June 12, 2007 3:53 AM
Subject: RE: [xsl] replace

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

-----Original Message-----
From: Charles Ohana [mailto:charles.ohana@xxxxxxxxxxxxxx]
Sent: 12 June 2007 03:28
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] replace

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="";
            xmlns:str=""; version="1.0"

<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>


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