Subject: Re: [xsl] Line end conversion. From: andrew.curry@xxxxxxxxxxxx Date: Tue, 2 Dec 2003 14:21:06 -0000 |
Something like this could be used <xsl:template name="replace"> <xsl:param name="string" select="." /> <xsl:choose> <xsl:when test="not($string)" /> <xsl:when test="contains($string, '|')"> <xsl:value-of select="substring-before($string, '|')" /> <br /> <xsl:call-template name="replace"> <xsl:with-param name="string" select="substring-after($string, '|')" /> </xsl:call-template> </xsl:when> <xsl:otherwise> <xsl:value-of select="$string" /> <br /> </xsl:otherwise> </xsl:choose> </xsl:template> <xsl:template match="string"> <tr> <td> <xsl:call-template name="replace" /> </td> </tr> </xsl:template> http://aspn.activestate.com/ASPN/Mail/Message/xsl-list/1393343 ----- Original Message ----- From: "Ben Trafford" <ben@xxxxxxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent: Tuesday, December 02, 2003 2:08 PM Subject: [xsl] Line end conversion. > > Hi! > > I have an odd little problem. I want to match all linefeeds with a given > element's content, and convert them to "\n". Does anybody know of a good > way to do so? > > In other words... > > <element>This is a phrase with a line end. > > It's very frustrating.</element> > > ...becomes... > > <element>This is a phrase with a line end.\n\nIt's very frustrating.</element> > > --->Ben > > > 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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