[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
Re: [xsl] outputing tags
Subject: Re: [xsl] outputing tags From: Sam Carleton <sam@xxxxxxxxxxxxxx> Date: Mon, 6 Jan 2003 11:42:54 -0500 |
On Mon, Jan 06, 2003 at 11:30:47AM -0500, Passin, Tom wrote: > [ Sam Carleton] > > > 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? > > > > It is all a matter of what you start with and what you want to end up > with. The stylesheet transforms the one into the other. > > So what do you want to end up with? You say you want a "line break" in > the XML, but XML inherently knows nothing about line breaks. Perhaps > you mean that you want to create html output that happens to be > well-formed xml as well? Then a <br/> element would just stay as is, if > there is one in the source. Otherwise, no one can answer you without > knowing how to determine where line breaks should occur. You are correct, I want the ,br/> in the html. > In general, if you just want to echo some existing element content > (complete with <br/> or other elements), you would use xsl:copy-of. If > you want to echo an element but change some of its child elements, use > an identity transform where you specify templates only for the elements > you want to change. The FAQs and archives contain examples. so that is what xsl:copy-of is for, I will look into that! Thanks a millon! Sam XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] outputing tags, Passin, Tom | Thread | [xsl] how to get an NCR in the outp, Tobias Reif |
Re: [xsl] [xml-doc] Newbie- Flow Di, Tobias Reif | Date | RE: [xsl] how to get < sign in outp, Passin, Tom |
Month |
Keywords