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Old November 28th, 2005, 03:31 PM
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Default Date format in XSLT

I have a date in the format mm/dd/yy
and i need this transformed to yyyymmdd

Can somebody suggest the method in xslt?
Old November 29th, 2005, 02:20 AM
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yes it is possbile. For that you have to write java extensions for xslt like

here the java code

package XSLT.Extension;

import java.io.*;
import java.util.*;
import java.text.*;

public class MyDateFormat {

    public static String getDateFormat(String dt)
        Date now = new Date();
        String outStr="";
                //===== converting String to date=========
                  String DATE_FORMAT = "dd/MM/yyyy";
                  SimpleDateFormat sdf = new SimpleDateFormat(DATE_FORMAT);

                  Date d=null;


                 //============ converting to yyyyMMdd format========
                SimpleDateFormat sdf1 = new SimpleDateFormat("yyyyMMdd");

          catch(Exception e)

        return outStr;

    public static String getOuts()

        return "TexTech international";

compile the java code and put the directory in classpath and
add the xmlns:dt="XSLT.Extension.MyDateFormat" to the XSL file
transform using the Saxon8 processor
you will get the output

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:dt="XSLT.Extension.MyDateFormat">

        <xsl:template match="/">

                            <xsl:value-of select="dt:getDateFormat('30/12/2004')"/>

Old November 29th, 2005, 04:10 AM
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You don't need extension functions to handle simple cases, just use something like

concat('20', substring($in, 7,2), '-', substring($in, 1,2), '-', substring($in,4,2))

Michael Kay
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

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