I am having the Date field as below
<Input>
<date>08/26/2020</date>
</Input>
i would need to parse it as like below
<date>2020-08-26</date>
I have tried using below xpath funtion in XSLT, which is not producing any result
xp20:format-dateTime(/Input/date,'[Y0001]-[M01]-[D01]')
Any help here??
Using replace you can reorder the components, if you want to create an XSLT/XPath xs:date, additionally use the constructor function:
<xsl:template match="date">
<xsl:copy>
<xsl:value-of select="xs:date(replace(., '([0-9]{2})/([0-9]{2})/([0-9]{4})', '$3-$1-$2'))"/>
</xsl:copy>
</xsl:template>
What you could do is, extract sub strings, and re-join them as needed.
<date><xsl:value-of select="concat(substring-after(substring-after(date/text(),'/'),'/'), '-', substring-before(date/text(),'/'), '-', substring-before(substring-after(date/text(),'/'),'/'))" /></date>
or better use a template for this:
<xsl:template name="format_date">
<xsl:param name="date" />
<xsl:value-of select="concat( substring($date, 7, 2),'.',substring($date, 5, 2), '.', substring($date, 1, 4), ', ', substring($date, 9, 2),':',substring($date, 11, 2),'h' )" />
</xsl:template>
and pass the date as param (with-param).
which'll give you the desired <date>2020-08-26</date>
