Re: [xsl] outputing tags

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?                                 

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