How to Convert date format in xslt (mm/dd/yyyy)

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Hi,

How to convert date format in xslt. I have created application as shown below

I used xml :
--------------

XML

<TRSummary>
  <TravellerRequest>
    <TravelRequestNumber>338</TravelRequestNumber>
    <OriginatorName>SureshVL RonVL</OriginatorName>
        <RequestDate>2015-01-06T17:51:01.67+05:30</RequestDate>
    <CostCenter>800</CostCenter>
    <ProjectManagerName>Prasannavl Athinavl Athinavl</ProjectManagerName>
    <JobNumber>1234</JobNumber>
    <ElementNumber>1234</ElementNumber>
    <Remarks>remarks</Remarks>
    <ExpectedTravelCost>1200</ExpectedTravelCost>
    <PMC>Industry international</PMC>
  </TravellerRequest>
</TRSummary>

My xslt file :
--------------

XML

<?xml version="1.0" encoding="utf-8"?>
<!-- Edited by XMLSpy� -->
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:template match="/">
     <xsl:for-each select="TRSummary">
      <html>
        <body>
          <table>
            <tr>
              <td>
            <xsl:value-of select="TravellerRequest/RequestDate"/>
              </td>
            </tr>
          </table>
        </body>
      </html>
    </xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

in html page getting out put like this :
---------------------------------------

Request Date: 2015-01-07T15:19:23.52+05:30

How to convert the above datetime format into mm/dd/yyyy.

i need output like this: 01/07/2015.

plz guide me.

Thanks,
Ram...

Posted 8-Jan-15 0:15am

vulisiramu

Updated 27-May-20 12:20pm

v2

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Zoltán Zörgő 11-Jan-15 6:52am

Any progress? If you have found my extensive answer of use, please accept it formally.

2 solutions

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**Zoltán Zörgő** · Answer 1 · 2015-01-08T01:44:00

Here you have more possibilities depending on XSLT version: http://blog.fpmurphy.com/2008/05/xslt-datetime-formatting.html[^]

Supposing following structure of the XML:

XML

<TRSummary>
  <TravellerRequest>
    <TravelRequestNumber>338</TravelRequestNumber>
    <OriginatorName>SureshVL RonVL</OriginatorName>
        <RequestDate>2015-01-06T17:51:01.67+05:30</RequestDate>
    <CostCenter>800</CostCenter>
    <ProjectManagerName>Prasannavl Athinavl Athinavl</ProjectManagerName>
    <JobNumber>1234</JobNumber>
    <ElementNumber>1234</ElementNumber>
    <Remarks>remarks</Remarks>
    <ExpectedTravelCost>1200</ExpectedTravelCost>
    <PMC>Industry international</PMC>
  </TravellerRequest>
  <TravellerRequest>
    <TravelRequestNumber>338</TravelRequestNumber>
    <OriginatorName>SureshVL RonVL</OriginatorName>
        <RequestDate>2015-01-06T17:51:01.67+05:30</RequestDate>
    <CostCenter>800</CostCenter>
    <ProjectManagerName>Prasannavl Athinavl Athinavl</ProjectManagerName>
    <JobNumber>1234</JobNumber>
    <ElementNumber>1234</ElementNumber>
    <Remarks>remarks</Remarks>
    <ExpectedTravelCost>1200</ExpectedTravelCost>
    <PMC>Industry international</PMC>
  </TravellerRequest>
</TRSummary>

This is one option:

XML

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

  <xsl:template match="/">
    <html>
       <body>
          <table border="1">
            <xsl:apply-templates select="TRSummary/TravellerRequest"/>
          </table>
       </body>
    </html>
  </xsl:template>

  <xsl:template match="RequestDate">
    <td>
      <xsl:value-of select="format-dateTime(.,'[M01]/[D01]/[Y0001]')" />
    </td>
  </xsl:template>

  <xsl:template match="TravellerRequest">
     <tr>
       <xsl:apply-templates select="RequestDate"/>
     </tr>
   </xsl:template>

</xsl:stylesheet>

[Update]
This is also working, but not so neat:

XML

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:template match="/">
      <html>
        <body>
          <table>
            <xsl:for-each select="TRSummary/TravellerRequest">
             <tr>
              <td>
                <xsl:value-of select="format-dateTime(RequestDate,'[M01]/[D01]/[Y0001]')" />
               </td>
             </tr>
            </xsl:for-each>
          </table>
        </body>
      </html>
  </xsl:template>
</xsl:stylesheet>

[Update 2]
This works with XSLT 1.0 too:

XML

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

  <xsl:template name="formatdate">
     <xsl:param name="DateTimeStr" />

     <xsl:variable name="datestr">
         <xsl:value-of select="substring-before($DateTimeStr,'T')" />
     </xsl:variable>

     <xsl:variable name="mm">
         <xsl:value-of select="substring($datestr,6,2)" />
     </xsl:variable>

     <xsl:variable name="dd">
        <xsl:value-of select="substring($datestr,9,2)" />
     </xsl:variable>

     <xsl:variable name="yyyy">
        <xsl:value-of select="substring($datestr,1,4)" />
     </xsl:variable>

     <xsl:value-of select="concat($mm,'/', $dd, '/', $yyyy)" />
  </xsl:template>

  <xsl:template match="/">
      <html>
        <body>
          <table border="1">
            <xsl:for-each select="TRSummary/TravellerRequest">
             <tr>
              <td>
                <xsl:call-template name="formatdate">
                    <xsl:with-param name="DateTimeStr" select="RequestDate"/>
                </xsl:call-template>
               </td>
             </tr>
            </xsl:for-each>
          </table>
        </body>
      </html>
  </xsl:template>
</xsl:stylesheet>

Praveen Kumar Upadhyay · Answer 2 · 2015-01-08T01:34:00

Solution 1

Check out below article. It might solve your issue.
http://www.sixtree.com.au/articles/2013/formatting-dates-and-times-using-xslt-2.0-and-xpath/[^]

Posted 8-Jan-15 1:34am

Praveen Kumar Upadhyay

Comments

vulisiramu 8-Jan-15 8:03am

I am not able to find the solution, plz could you help me.

vulisiramu 8-Jan-15 23:58pm

Hi praveen, Plz provide answer to above question?

Praveen Kumar Upadhyay 9-Jan-15 1:15am

format-date() is not working for you?

Zoltán Zörgő 9-Jan-15 2:39am

And what about the solution I had given? It is tested ad working... please note that your approach won't be valid if more than one requests are in the xml. See update for a solution more like your approach - but please note the reordering of the elements in the xslt!

Praveen Kumar Upadhyay 9-Jan-15 7:30am

Okay. Then OP is free to accept your solution.

vulisiramu 9-Jan-15 7:36am

ok, plz guid me how to resolve this error:
i am getting below error
Exception Details: System.Xml.Xsl.XsltException: 'format-dateTime()' is an unknown XSLT function.

How to Convert date format in xslt (mm/dd/yyyy)

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How to Convert date format in xslt (mm/dd/yyyy)

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