Need to format the date as per the input xml file - xslt

Question

Need to change the date format from 123121(mmddyy) to 12/31/2021(mm/dd/yyyy)
Input XML I'm having:
<section>
<Plans>
<Date>123121</Date>
</Plans>
</section>
<p>date: <keyword keyref="Cost:Date:date4"/></p>
XSL I have tried:
<xsl:template match="keyword[contains(#keyref, 'Cost:Date:date4')]">
<xsl:param name="section" as="element()" tunnel="yes">
<empty/>
</xsl:param>
<keyword keyref="Cost:Date:date4">
<xsl:call-template name="format_variable">
<xsl:with-param name="cur_keyref" select="#keyref"/>
<xsl:with-param name="cur_value"
select="$section//Plans/Date"/>
<xsl:with-param name="cur_format" select="'date4'"/>
</xsl:call-template>
</keyword>
</xsl:template>
<xsl:template name="format_variable">
<xsl:param name="cur_keyref" as="xs:string">MISSING</xsl:param>
<xsl:param name="cur_value" as="xs:string">MISSING</xsl:param>
<xsl:param name="cur_format" as="xs:string">MISSING</xsl:param>
<xsl:choose>
<xsl:when test="$cur_format = 'date4'">
<xsl:value-of select="format-dateTime($cur_value, '[M01]/[D01]/[Y0001]')"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of
select="concat('MISSING_FORMAT_', $cur_keyref, '_', $cur_value, '_[', $cur_format, ']')"
/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
When I'm trying the above code its showing the below error:
Invalid dateTime value "123121" (Too short)
Please somebody help me out get the 12/31/2021(mm/dd/yyyy) value on the output.

Answer 1

You are trying to use the format-dateTime() function on something that is not a valid dateTime.
What can't you do simply:
<xsl:value-of select="replace(Date, '(.{2})(.{2})(.{2})', '$1/$2/20$3')"/>
Note that this is assuming all your dates will be in the 21st century. Otherwise you would need additional logic to identify the century.