Subject: RE: [xsl] Can I take only parts from node value? From: "jp" <jp@xxxxxxxxxxxxxxx> Date: Tue, 19 Feb 2002 18:27:56 +0100 |
Yes, I got it working! Thanks for help :) XML: <calendar date="200202180800" /> XSL: <xsl:variable name = "year" select = "substring(@date,0,5)" /> <xsl:variable name = "month" select = "substring(@date,5,2)" /> <xsl:variable name = "day" select = "substring(@date,7,2)" /> <xsl:variable name = "hour" select = "substring(@date,9,2)" /> <xsl:variable name = "minute" select = "substring(@date,11)" /> <xsl:value-of select="$year"/> <xsl:value-of select="$month"/> <xsl:value-of select="$day"/> <xsl:value-of select="$hour"/> <xsl:value-of select="$minute"/> Regs, -JP -----Original Message----- From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of Michael Kay Sent: 19. helmikuuta 2002 18:06 To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: RE: [xsl] Can I take only parts from node value? > 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? With the substring() function. Michael Kay Software AG home: Michael.H.Kay@xxxxxxxxxxxx work: Michael.Kay@xxxxxxxxxxxxxx XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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