Need to reduce the 20 days prior from the input xml - datetime

Question

Need to get the 20 days prior from the input xml element.
Input xml I'm having(year-month-day):
<section>
<Plans>
<Date>2022-01-01</Date>
</Plans>
</section>
<p>20 days back date: <keyword keyref="Cost:Date:date4"/></p>
XSL I'm using
<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:variable name="format" select="replace($cur_value, '(.{4})-(.{2})-(.{2})', '$2/$3/$1')"/>
<xsl:value-of select="$format - xs:dateTime('P20D')"/>
</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 "P20D" (Non-numeric year component)
Expected output is:
20 days back date: 12/12/2021

Answer 1

FWIW, I believe you are grossly overcomplicating this. The date in the input XML is in YYYY-MM-DD format; all you need to do is subtract a duration of 20 days from the given date - see demo here: https://xsltfiddle.liberty-development.net/6qaCymJ

Answer 2

I think you want to subtract a dayTimeDuration with e.g. <xsl:value-of select="$format - xs:dayTimeDuration('P20D')"/>.