Here's one way:
var xml:XML = my_xml
row id=a parent=a /
row id=b parent=a /
row id=c parent=a /
row id=d parent=d /
row id=e parent=d /
On 16/10/2011 10:06, Kenneth Kawamoto wrote:
Here's one way:
var xml:XML = my_xml
row id=a parent=a /
row id=b parent=a /
row id=c parent=a /
row id=d parent=d /
row id=e parent=d /
row id=f parent=d /
/my_xml;
var ids:Array = [a, d];
for each (var s:String in ids){
xml.row.(@id ==
On 16/10/2011 13:01, Paul Andrews wrote:
On 16/10/2011 10:06, Kenneth Kawamoto wrote:
Here's one way:
var xml:XML = my_xml
row id=a parent=a /
row id=b parent=a /
row id=c parent=a /
row id=d parent=d /
row id=e parent=d /
row id=f parent=d /
/my_xml;
var ids:Array = [a, d];
for each (var
It's getting rather interesting :D
var xml:XML = my_xml
row id=a parent=a /
row id=b parent=a /
row id=c parent=a /
row id=d parent=d /
row id=e parent=d /
row id=f parent=d /
On 16/10/2011 14:16, Kenneth Kawamoto wrote:
It's getting rather interesting :D
var xml:XML = my_xml
row id=a parent=a /
row id=b parent=a /
row id=c parent=a /
row id=d parent=d /
row id=e parent=d /
row id=f parent=d /
row id=g parent=e /
row id=h parent=e /
row id=i parent=h /
/my_xml;
On 12/10/2011 21:49, Kenneth Kawamoto wrote:
Not sure if this is the best way but you can do:
var xml:XML = my_xml
row id=a /
row id=b /
row id=c /
/my_xml;
xml.row.(@id == a).children = xml.row.(@id != a);
xml.setChildren(xml.row.(@id == a));
trace(xml);
// trace
my_xml
row id=a
row id=b/
On 11/10/2011 17:13, Glen Pike wrote:
Hi,
You would have to:
1 Create a new XML node,
myNewNode = my_xml/
2. Select nodes from the existing XML where id != a
var nodes:XMLList = my_xml.child(row).attribute(id != a);
3. Select nodes from existing XML where id = a
Not sure if this is the best way but you can do:
var xml:XML = my_xml
row id=a /
row id=b /
row id=c /
/my_xml;
xml.row.(@id == a).children = xml.row.(@id != a);
xml.setChildren(xml.row.(@id == a));
row id=c /
/row
And
row id=c /
/row
Are not valid XML. Therefore, I'm not clear on what you want to do. :)
Jason Merrill
Instructional Technology Architect II
Bank of America Global Learning
___
-Original Message-
From:
On 11/10/2011 16:33, Merrill, Jason wrote:
row id=c /
/row
And
row id=c /
/row
Are not valid XML. Therefore, I'm not clear on what you want to do. :)
Hmm..
myXMLNode =
my_xml
row id=a /
row id=b /
row id=c /
/my_xml;
into:
my_xml
row id=a
row id=b /
row id=c /
/row
/my_xml
I didn't
Hi,
You would have to:
1 Create a new XML node,
myNewNode = my_xml/
2. Select nodes from the existing XML where id != a
var nodes:XMLList = my_xml.child(row).attribute(id != a);
3. Select nodes from existing XML where id = a
var parentNodes:XMLList =
Subject: Re: [Flashcoders] E4X XML manipulation
On 11/10/2011 16:33, Merrill, Jason wrote:
row id=c /
/row
And
row id=c /
/row
Are not valid XML. Therefore, I'm not clear on what you want to do. :)
Hmm..
myXMLNode =
my_xml
row id=a /
row id=b /
row id=c /
/my_xml;
into:
my_xml
___
-Original Message-
From: flashcoders-boun...@chattyfig.figleaf.com
[mailto:flashcoders-boun...@chattyfig.figleaf.com] On Behalf Of Paul Andrews
Sent: Tuesday, October 11, 2011 11:54 AM
To: Flash Coders List
Subject: Re: [Flashcoders] E4X XML manipulation
On 11/10/2011 16:33
...@chattyfig.figleaf.com] On Behalf Of Paul Andrews
Sent: Tuesday, October 11, 2011 12:30 PM
To: flashcoders@chattyfig.figleaf.com
Subject: Re: [Flashcoders] E4X XML manipulation
On 11/10/2011 17:21, Merrill, Jason wrote:
This, what you are saying you want to turn it into:
row id=c /
/row
That actually
On 11/10/2011 17:38, Merrill, Jason wrote:
Just trying to help. :)
Yes, I appreciate it. Thank you.
I don't know if this helps what you're trying to do, but I have found it
much easier to parse XML data into value objects and vectors, then doing the
data manipulations from there, rather
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