2010-07-11 6 views

Répondre

26

réponse courte, oui:

var xnList:XMLList = xml.Names.Name.(@type == "M"); 

version plus longue:

var xml:XML = <Root> 
     <Names> 
      <Name type="M" value="John Doe" /> 
      <Name type="F" value="Jane Doe" /> 
      <Name type="M" value="John Hancock" /> 
     </Names> 
     <Other /> 
     </Root> 

var xnList:XMLList = xml.Names.Name.(@type == "M"); 

//test 
for each (var xnNode:XML in xnList) trace(xnNode.toXMLString()) 

Il y a une assez bonne E4X tutorial on the Yahoo Developer Network.

HTH

+0

+1 bonne réponse. – gMale

+0

+1 pour la réponse – o15a3d4l11s2

+0

Pouvez-vous faire des jokers? // * – FlavorScape

Questions connexes