RE: [xsl] Unable to replace "/"

Subject: RE: [xsl] Unable to replace "/"
From: "V.Ramkumar" <v.ramkumar@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 23 Dec 2008 17:14:38 +0530
I have applied replace method in copy-of select. Now I got it, please excuse
me

Regards,
Ramkumar


-----Original Message-----
From: V.Ramkumar [mailto:v.ramkumar@xxxxxxxxxxxxxxxxxxxxxx] 
Sent: Tuesday, December 23, 2008 4:59 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] Unable to replace "/"

Hi list,

Unable to replace /. Please see below and provide solutions.

XSL:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">
<xsl:output omit-xml-declaration="yes"/>

    <xsl:template match="/">
        <xsl:apply-templates/>
    </xsl:template>

    <xsl:template match="root">
        <xsl:for-each select="date">
            <xsl:sort
select="replace(concat(substring(.,7,2),substring(.,4,2),substring(.,1,2)),'
/','-')" order="ascending" data-type="text" /><xsl:copy-of
select="."/><xsl:text>&#xa;</xsl:text>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

XML: 
<?xml version="1.0" encoding="UTF-8"?>
<root>
    <date>07/12/07</date>
    <date>06/11/05</date>
    <date>06/12/06</date>
    <date>05/12/09</date>
    <date>05/02/08</date>
</root>

Expected O/P:

<date>06-11-05</date>
<date>06-12-06</date>
<date>07-12-07</date>
<date>05-02-08</date>
<date>05-12-09</date>

Regards,
Ramkumar.

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