[xsl] Can I take only parts from node value?

<-prev [Thread] next->	<-prev [Date] next->
Month Index \| List Home

[xsl] Can I take only parts from node value?

Subject: [xsl] Can I take only parts from node value?
From: "jp" <jp@xxxxxxxxxxxxxxx>
Date: Tue, 19 Feb 2002 16:58:28 +0100

Hi,

If I have XML-tag like this:

<today date="200202190800" />

and I want output:

Time is 08:00
Date is 19.02 
Year is 2002

How can I do this with XSL/T? Or should I change XML output?

Regs,
	-JP


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list

Current Thread
[xsl] regd: XSLT in SAXON laksh vr - Tue, 19 Feb 2002 06:02:18 -0800 (PST) Michael Kay - Tue, 19 Feb 2002 14:25:00 -0000 jp - Tue, 19 Feb 2002 16:58:28 +0100 <= Robert Koberg - Tue, 19 Feb 2002 08:07:32 -0800 Oleg Tkachenko - Tue, 19 Feb 2002 18:21:41 +0200 Michael Kay - Tue, 19 Feb 2002 17:05:31 -0000 jp - Tue, 19 Feb 2002 18:27:56 +0100

<- Previous	Index	Next ->
RE: [xsl] regd: XSLT in SAXON, Michael Kay	Thread	Re: [xsl] Can I take only parts fro, Robert Koberg
[xsl] How to select a numbered grou, philip . borg	Date	Re: [xsl] Q on incremental processi, Tobias Kiesling
	Month