[Lift] Re: Concatenating lists of nodes

2008-11-13 Thread Jorge Ortiz
A NodeSeq is really just a Seq[Node]. And a List[Node] is also a Seq[Node].
Hence a List[Node] is a NodeSeq.
scala import scala.xml.NodeSeq
import scala.xml.NodeSeq
scala val n: NodeSeq = List(spanA/span, spanB/span, spanC/span)
n: scala.xml.NodeSeq = spanA/spanspanB/spanspanC/span

--j

On Thu, Nov 13, 2008 at 10:14 AM, Charles F. Munat [EMAIL PROTECTED] wrote:


 If I have a list thus:

 List(spanA/span,spanB/span,spanC/span)


 How can a get a NodeSeq thus from this:

 spanA/spanspanB/spanspanC/span


 I'm sure this is drop-dead simple, but it's still not obvious to me...

 Thanks,

 Chas.

 


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[Lift] Re: Concatenating lists of nodes

2008-11-13 Thread Charles F. Munat

Sigh... I knew it had to be drop-dead simple. Not sure how I kept 
getting List(span...) output instead of span.../span out. Works 
now. Thanks.

Chas.

Jorge Ortiz wrote:
 A NodeSeq is really just a Seq[Node]. And a List[Node] is also a 
 Seq[Node]. Hence a List[Node] is a NodeSeq.
 
 scala import scala.xml.NodeSeq
 import scala.xml.NodeSeq
 scala val n: NodeSeq = List(spanA/span, spanB/span, spanC/span)
 n: scala.xml.NodeSeq = spanA/spanspanB/spanspanC/span
 
 --j
 
 On Thu, Nov 13, 2008 at 10:14 AM, Charles F. Munat [EMAIL PROTECTED] 
 mailto:[EMAIL PROTECTED] wrote:
 
 
 If I have a list thus:
 
 List(spanA/span,spanB/span,spanC/span)
 
 
 How can a get a NodeSeq thus from this:
 
 spanA/spanspanB/spanspanC/span
 
 
 I'm sure this is drop-dead simple, but it's still not obvious to me...
 
 Thanks,
 
 Chas.
 
 
 
 
  

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[Lift] Re: Concatenating lists of nodes

2008-11-13 Thread Matt Harrington

On Thu, Nov 13, 2008 at 10:32 AM, Matt Harrington [EMAIL PROTECTED] wrote:

 Take a look at List.mkString().  It inherits from Iterable.

 ---Matt

Erm, I'll retract this.  :)

---Matt

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