Subject: Re: [xsl] outputing tags From: Sam Carleton <sam@xxxxxxxxxxxxxx> Date: Sat, 4 Jan 2003 20:36:35 -0500 |
On Sat, Jan 04, 2003 at 06:43:46PM -0500, cknell@xxxxxxxxxx wrote: > The XML sample you listed, is it the source which you are > trying to transform, or is the the output you desire to > receive? In either case, send us a sample of the other and > it will be easier to help. Sorry about that. The XML is the source. I would like the output to look something like this: <b>This is to be displayed<br/>This is a new line<br/>And this is on a third line</b> Someone else posted with the obvious correct answer. I thought I had tried it, but guess I hadn't. > Rap yourself on the knuckles with a ruler! We do not output > tags, we output a node tree comprised of elements (which are > bounded by opening and closing tags). Ok, then how do I do this: I have my xml file with all the info I need in it. There is one element named <quote> which contains quotes. Some quotes need to have line break ( <br>'s in HTML). How do I denote a line break in the XML and then how do I transform it? Sam > -----Original Message----- > From: Sam Carleton <sam@xxxxxxxxxxxxxx> > Sent: Sat, 4 Jan 2003 14:35:39 -0500 > To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] outputing tags > > Folks, > > How do I output some tags? The xml looks like this: > > <sometag>This is to be displayed<br/>This is a new > line<br/>And this is on a thrid line</sometag> > > Then I have a template like this: > > <xsl:template match="sometag"> > <b><xsl:apply-templates/></b> > </xsl:template> > > How do I write a template that will output both of the > <br/>'s? so that it is displayed the right way? XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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