i have an XML file which has a date in this format
<ROWSET>
<ROW>
<name> John </name>
<order_date> 10/1/2008</order_date>
<amount> 122 </amount>
</ROW>
<ROW>
<name> Jack </name>
<order_date> 12/6/2008</order_date><amount> 245 </amount>
</ROW>
</ROWSET>

In xsl when i do this

<xsl:stylesheet version="1.0"
xmlns="urn:schemas-microsoft-com:office:spreadsheet"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet"
xmlns:o="urn:schemas-microsoft-com:office:office"
xmlns:x="urn:schemas-microsoft-com:office:excel">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/">
<Workbook>
<Worksheet ss:Name="Biogen_Reverse_Payment_CT">
<Table x:FullColumns="1" x:FullRows="1">
<xsl:for-each select="ROWSET/ROW">
<Row>

<Cell><Data ss:Type="String">
<xsl:value-of select="CURRENCY_CODE"/>
</Data></Cell>

<Cell> <Data ss:Type="DateTime">
<xsl:value-of select="CHECK_DATE"/>
</Data> </Cell>

<Cell><Data ss:Type="Number">
<xsl:value-of select="AMOUNT"/>
</Data></Cell>
</Row>
</xsl:for-each>
</Table>
</Worksheet>
</Workbook>
</xsl:template>
</xsl:stylesheet>

In the resulting output Excel file from this I cannot open the excel fuile as DateTime format is like this "12/6/2008" but in xsl it wants like this "2007-10-18T00:00:00.000"

How do i fix this so i can open the file in excel.

Both your example dates are ambiguous, so I'm not sure when you say "12/6/2008" whether you mean 12th June or 6th December.

Your best bet is to avoid such date formats in your XML, and use the international standard format YYYY-MM-DD. If you can't do that then you will have to convert it "by hand", which you can do fairly easily in XSLT 2.0 using regular expressions, or rather more tediously in XSLT 1.0 using string functions such as substring-before, etc.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

How do i convert from this format mm/dd/yyyy to this format

"2007-10-18T00:00:00.000" using XSL. can you pls tell me any function or example to do it.

I am getting the XML file from a database. so i can get only in this format mm/dd/yyyy

XSLT 1.0 or 2.0?

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

XSLT 1.0

is there any function to convert mm/dd/yyyy or from YYYY-MM-DD HH24:MI:SS to 2007-10-18T00:00:00.000

I appreciate ur help.
Thanks

Extract the components into variables, for example the year part is substring-after(substring-after(., '/'), '/'). Use format-number() to force the month and day to two digits. Then use concat() to reassemble the date in the required format.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

can you please provide an example code so i can modify it accordingly.

Please try to write the code yourself. If you can't make it work, show us your attempts and we'll tell you where you went wrong. It's a tedious problem but not difficult - I'm happy to help people with genuine problems understanding the specs, but I'm not going to sit here doing your work for you.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference