how to change Dateformate in XSLT?

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hello friends,

i have below code of xslt. how can i change the date formate in xslt.?

now it's returning in "DD/MON/YY" formate.but i want it in "DD/MM/YYYY" formate.

i have tried code which is in commented part...
See the below code....

HTML

<xsl:element name="{$rowName}">
    <xsl:for-each select="$fieldset[@name = $rowField]/s:AttributeType">
        <xsl:variable name="fieldname" select="@name" />
        <xsl:variable name="dataType" select="$fieldset[@name = $rowField]/s:AttributeType[@name=$fieldname]/s:datatype/@dt:type" />
        <xsl:variable name="fieldvalue">
            <xsl:choose>
               <xsl:when test="$dataType = 'dateTime'">

                 <xsl:value-of select="substring($row/@*[name()=$fieldname], '1', '2')"/>

        <!-- i have tried below code to get date in  MM formate.. but... :( -->

                <!--<xsl:variable name="mm">
                    <xsl:value-of select="substring($row/@*[name()=$fieldname],4,2)" />
                </xsl:variable>
                <xsl:value-of select="$mm"/> -->

               </xsl:when>
               <xsl:otherwise>
                 <xsl:value-of select="$row/@*[name()=$fieldname]"/>
               </xsl:otherwise>
            </xsl:choose>
        </xsl:variable>
        <xsl:element name="{$fieldname}"><xsl:value-of select="$fieldvalue" /></xsl:element>
        <!--xsl:attribute name="{$fieldname}"><xsl:value-of select="$fieldvalue" /></xsl:attribute-->
    </xsl:for-each>
    <!--RECURSIVELY CALL THIS TEMPLATE IF IT CONTAINS ANY CHILD ELEMENTS-->
    <xsl:apply-templates select="*" />
</xsl:element>

Thanx in advance....

Posted 16-Jul-12 19:44pm

ssd_coolguy

Updated 16-Jul-12 21:06pm

v3

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Prasad_Kulkarni · Answer 1 · 2012-07-17T19:14:00

Try this way:

C#

string ms:format-date(string datetime, string format [,locale])

This will work for you:

C#

<xsl:value-of select="ms:format-date(., 'dd/MM/yyyy')" xmlns:xsl="#unknown" />

Refer DateFormats: http://msdn.microsoft.com/en-us/library/ms256099.aspx[^]

Have a look on example:
XSLT DateTime Formatting[^]

And a similar example to your question:
xsl format date dd/MM/yyyy[^]

Also check this one:
Formatting Dates and Times[^]
Here three functions are provided to represent dates and times as a string, using the conventions of a selected calendar, language, and country.