XSLT convert date format

Question

I'm having an issue where I'm getting an XML file, and from the samples that I get the format of date that I'm getting is mm/dd/yyyy and sometimes it is m/d/yyyy. My task is to convert this to another XML file where the schema only accepts yyyy-mm-dd. I'm limited to using XSLT 1.0/XPATH 1.0. How can I do this?

Solution 1

Your question is a little vague. It needs some sample input and expected output. But any way, here is an answer to best guess at what you want.

Given this input:

<?xml version="1.0"?>
<dates>
  <date>11/12/2012</date>
  <date>3/4/2011</date>
</dates>

... transformed by this style-sheet ...

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>

      <xsl:template match="dates">
        <xsl:copy>
        <xsl:apply-templates select="*" />
        </xsl:copy>
      </xsl:template>

      <xsl:template match="date">
        <xsl:copy>
        <xsl:value-of select=" 
           concat(
           substring-after(substring-after(.,'/'),'/') , '-',
           format-number( number( substring-before(.,'/')), '00') , '-',
           format-number( substring-before(substring-after(.,'/'),'/'), '00') 
           )
          " />
        </xsl:copy>
      </xsl:template>

</xsl:stylesheet>

... will produce this desired output ...

<dates>
 <date>2012-11-12</date>
 <date>2011-03-04</date>
</dates>

Please tick the answer if it is correct. I verified this solution on http://www.purplegene.com/static/transform.html

Solution 2

My XSLT shown below should work. It does not use hard-coded lengths or indexes, but instead splits the date string on '/' to work out where the day, month and year components are.

XSLT:

<?xml version="1.0" encoding="utf-8"?>

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" />

    <xsl:template match="/">
        <xsl:apply-templates />
    </xsl:template>

        <!-- match the date and invoke formatDate to format it -->
    <xsl:template match="date">
        <xsl:element name="date">
            <xsl:call-template name="formatDate">
                <xsl:with-param name="dateParam" select="." />
            </xsl:call-template>
        </xsl:element>
    </xsl:template>

    <xsl:template name="formatDate">
        <xsl:param name="dateParam" />
        <!-- input format mm/dd/yyyy or m/d/yyyy -->
        <!-- output format yyyy-mm-dd -->


        <!-- parse out the day, month and year -->
        <xsl:variable name="month" select="substring-before($dateParam,'/')" />
        <xsl:variable name="day" select="substring-before(substring-after($dateParam,'/'),'/')" />
        <xsl:variable name="year" select="substring-after(substring-after($dateParam,'/'),'/')" />

        <!-- now print them out. Pad with 0 where necessary. -->
        <xsl:value-of select="$year" />
        <xsl:value-of select="'-'" />
        <xsl:if test="string-length($month) = 1">
            <xsl:value-of select="'0'" />
        </xsl:if>
        <xsl:value-of select="$month" />
        <xsl:value-of select="'-'" />
        <xsl:if test="string-length($day) = 1">
            <xsl:value-of select="'0'" />
        </xsl:if>
        <xsl:value-of select="$day" />
    </xsl:template>

    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()" />
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

Input:

<input>
    <date>04/30/2012</date>
    <date>4/1/2012</date>
</input>

Output:

<?xml version="1.0" encoding="UTF-8"?>
<input>
    <date>2012-04-30</date>
    <date>2012-04-01</date>
</input>

Demo: http://www.xsltcake.com/slices/kBveVQ

Comments

[Sean B. Durkin]

Dogbane's solution is pretty similiar to mine, but he beat me to the punch. Still, I think my solution is briefer and reads better. McLaren's solution won't work on dates like 1/12/2012 which has string length 9.

[Michael Kay]

Please don't use <xsl:variable><xsl:value-of select="xxx"/></xsl:variable> when you could use <xsl:variable select="xxx"/>. It's verbose and inefficient!

[Mikro Koder]

Hello team! See below , exposing format-dateTime: stackoverflow.com/a/16892467/2816548