Ah, this is exactly what I'm looking for. Thank you both!
-KP
On Sep 1, 10:20 am, David Pollak
wrote:
> If you have a String that you know is valid XHTML but don't want to go
> through the parsing/unparsing phase:
>
> import scala.xml._
>
> val s: String =
>
> {Unparsed(s)}
>
> The String
If you have a String that you know is valid XHTML but don't want to go
through the parsing/unparsing phase:
import scala.xml._
val s: String =
{Unparsed(s)}
The String will be included in the output stream unchanged.
On Mon, Aug 31, 2009 at 11:49 PM, marius d. wrote:
>
>
>
> {(str.spl
{(str.split("\n").map(x => {x}))}
{x} is an expression that returns a String. Assuming that x is a well-
formed XML node you could try something like:
{(str.split("\n").map(x => {XML.loadString(x)}))}
Br's,
Marius
On Sep 1, 7:24 am, KP wrote:
> Hi all.
>
> I have some text I read in
