Re: Background groups

2013-02-08 Thread Robert Sneidar
Ah! I thought you meant place by script. Bob On Feb 8, 2013, at 1:17 PM, Peter Haworth wrote: > Thanks Devin, perfect write up as usual. Even answered my reply to Scott > about how to place a group back on a card after it's been removed from all > cards (Place Group on the Object menu). > >

Re: Background groups

2013-02-08 Thread Robert Sneidar
You've got it. Refer to it by name or ID, assuming your names are unique, or ID. I'm a staunch believer in every object having a unique name anyway, so I prefix all my objects with a 2 or 3 character code like bg, btn, fld, mnu etc. Sorts nicely in lists too. Bob On Feb 8, 2013, at 1:13 PM,

Re: Background groups

2013-02-08 Thread Peter Haworth
Thanks Devin, perfect write up as usual. Even answered my reply to Scott about how to place a group back on a card after it's been removed from all cards (Place Group on the Object menu). I think I'm good to go now. On Fri, Feb 8, 2013 at 11:49 AM, Devin Asay wrote: > http://livecode.byu.edu/

Re: Background groups

2013-02-08 Thread Peter Haworth
Thanks Scott, think I have that straight. On the remove command, I think I'm right in saying that if I remove a group from a card and it's the last instance of the group in the stack, it is still in the stack even though it's not any card. If I subsequently want to put that group back on a card h

Re: Background groups

2013-02-08 Thread Devin Asay
Hi Pete! On Feb 8, 2013, at 12:27 PM, Peter Haworth wrote: > Trying to avoid any gotchas before embarking on some work involving > background groups and there's precious little information about them. This is how I introduce groups to my students. (Apologies if it's too basic. And I can see now

Re: Background groups

2013-02-08 Thread Scott Rossi
Hi Pete: "Place" adds an instance of an existing group to a card. Copy & paste creates a new group object. Deleting a group physically deletes it from the stack. Removing a group removes the instance of the group from the card, but doesn't delete the group from the stack. Regards, Scott Rossi