Subject: [xsl] Re: date format using xslt 1.0 From: Jacob L <arkle.ash@xxxxxxxxx> Date: Fri, 15 Jun 2012 13:09:18 -0400 |
Hi, I am using this xslt shared by Mr.David.Can anyone help me with the code to convert the time also.I previously sent message that it is not important,but now the req. has changed.Thanks. <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="date"> <xsl:variable name="a" select="substring-after(.,', ')"/> <xsl:variable name="d" select="substring-before($a,' ')"/> <xsl:variable name="b" select="substring-after($a,' ')"/> <xsl:value-of select="concat( substring-before(substring-after($b,' '),' '), format-number(string-length(substring-before('xxxJanFebMarAprMayJunJulAugSepOctNovDec',substring-before($b,' '))) div 3,'00'), format-number($d,'00') )"/> </xsl:template> </xsl:stylesheet> On Tue, Jun 12, 2012 at 1:47 PM, Jacob L <arkle.ash@xxxxxxxxx> wrote: > Hi, > > > I am trying to write an XSLT for formatting date.I am unable to find > any such previous posting,so sending it to the list.Here is my > requirement. > > Input format: > > <date>Mon, 11 Jun 2012 17:29:42 +0000 </date> > > After the transform,I want output as:- > > <date>20120611</date> which is YYYYMMDDkkmmss > > Time at the end is not important.I am using XSLT 1.0 and SAX parser. > > Thanks a lot!
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