It's best to avoid date formats such as d/m/y or m/d/y - better to use the
ISO date format yyyy-mm-dd. This is supported directly in XSLT 2.0 via the
data type xs:date. If you're stuck with XSLT 1.0 and/or d/m/y format dates,
you will have to rearrange them into y/m/d format using a combination of
concat() and substring() operations, you can then convert 21/02/05 to
20050221, which you can then sort directly either as a number or as a
string. So it's something like
<xsl:sort select="concat(substring(.,7,2), substring(.,4,2),
substring(.,1,2)"/>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: V.Ramkumar [mailto:v(_dot_)ramkumar(_at_)macmillansolutions(_dot_)com]
Sent: 23 December 2008 09:53
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Sorting date format
Hi List,
How to sort the date field. Please see below and provide solution.
XML:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<date>07/12/07</date>
<date>11/12/06</date>
<date>10/12/05</date>
<date>21/02/05</date>
</root>
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="." order="ascending" data-type="text"
/><xsl:copy-of select="."/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Required Output:
<date>21/02/05</date>
<date>10/12/05</date>
<date>11/12/06</date>
<date>07/12/07</date>
Regards,
Ramkumar
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