Not sure I understand the question, but the source of an AC is never modified, so you can always make a copy of AC.source via AC.source.concat() to get a fresh unfiltered/unsorted copy of the data even if the AC currently is sorted/filtered.
If the AC is sorted/filtered, you can copy that sorted/filtered version by calling toArray() ________________________________ From: flexcoders@yahoogroups.com [mailto:[EMAIL PROTECTED] On Behalf Of Dominic Pazula Sent: Friday, April 11, 2008 11:04 AM To: flexcoders@yahoogroups.com Subject: [flexcoders] Re: Copying View of an ArrayCollection I am not aware of a way to do what you want. I agree resorting/filtering is inefficient. Could you programmatically check for when that condition occurs and when it does just set the destination equal to the source? If you want to keep the destination around, then maybe set whatever is using the destination to use the source instead? --- In flexcoders@yahoogroups.com <mailto:flexcoders%40yahoogroups.com> , "m.ar80" <[EMAIL PROTECTED]> wrote: > > --- In flexcoders@yahoogroups.com <mailto:flexcoders%40yahoogroups.com> , "Dominic Pazula" <dompazz@> wrote: > > > > Are you looking for the two ArrayCollections to always have the same > > filter and sort, or only occasionally? > > > > Hi Dominic. Thank you for your response. > > The two array collections will share the same source but not > necessarily have the same filter and sort throughout their lifetimes. > Only occasionally, and this is when I would like to copy source > collection's view directly rather than re-filter and re-sort items in > the destination collection. >
