to be safe (assuming you don't nest) channels..channel[...@name
its the '.@' that is missing in your code: you can also do... myXMLroot..(@name == "bob") ; // doesn't matter how nested the node with "bob" as an attribute is. On Sun, Mar 29, 2009 at 8:16 AM, Justin Nichols <[email protected]>wrote: > See if this helps: > > var xml:XML = > <rootTag> > <channel name="a"/> > <channel name="b"/> > <channel name="c"/> > </rootTag>; > > var list:XMLList = xml.channel; > > var firstChannelName:String = list[...@name; // may need to toString() or > cast as String > > Thanks, > > Justin Nichols > > > > On Mar 29, 2009, at 8:08 AM, Atlanta Geek wrote: > > So I have an XMLList (named channels) that looks like this in xml >> >> <channel name="a"/> >> <channel name="b"/> >> <channel name="c"/> >> >> It is the data source for a combobox that is an item editor in a datagrid. >> >> So how do I access the name attribute of the second channel. >> channels[1].name does not give me the name value. Any suggestions? >> >> -- >> http://www.atlantageek.com >> >> >> ------------------------------------------------------------- >> To unsubscribe from this list, simply email the list with unsubscribe in >> the subject line >> >> For more info, see http://www.affug.com >> Archive @ http://www.mail-archive.com/discussion%40affug.com/ >> List hosted by http://www.fusionlink.com >> ------------------------------------------------------------- >> >> >> > > > ------------------------------------------------------------- > To unsubscribe from this list, simply email the list with unsubscribe in > the subject line > > For more info, see http://www.affug.com > Archive @ http://www.mail-archive.com/discussion%40affug.com/ > List hosted by http://www.fusionlink.com > ------------------------------------------------------------- > > > -- Darin Kohles RIA Developer
