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Differentiating Properties by their Property Groups

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Remi Monedi
Differentiating Properties by their Property Groups
on Oct 23, 2018 at 11:16:07 pm

Hi,

Is it possible to differentiate 2 properties with the same name (e.g. "Slider") without using their index?

For instance, each "Slider" properties are in a different Property Group : "Group 1" & "Group 2" in an effect "EFFECT".


If I'd like to link the "Slider" from "Group 2", can I differentiate it with something like that? (this won't work)
thisComp.layer("LAYER").effect("EFFECT")("Group 2")("Slider")


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Alex Printz
Re: Differentiating Properties by their Property Groups
on Oct 24, 2018 at 2:18:08 pm

I don't think so; effect elements stack as layers, the sub-groups inside effects are just peudo-groups that are for usability only but are not used in actual referencing.

Your best bet would be referencing indexes and relative use, or using a name structure in your effect that you can add/subtract strings from to call sections appropriately.

Alex Printz
Mograph Designer


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Remi Monedi
Re: Differentiating Properties by their Property Groups
on Oct 24, 2018 at 3:24:00 pm

Thank you for your answer Alex. I wasn't sure about the referencing of Property Group names.
So I guess nothing can be made using a "propertyGroup(x)" expression?


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Dan Ebberts
Re: Differentiating Properties by their Property Groups
on Oct 24, 2018 at 3:55:01 pm

propertyGroup() is useful in navigating a property's hierarchy, but it I don't think it helps you in this case because Group 1, Group 2, and both Slider properties are all at the same level in the hierarchy. They are all direct children of the effect itself, which makes them siblings. The appearance of child/parent relationship in the UI is just cosmetic.

Dan



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Remi Monedi
Re: Differentiating Properties by their Property Groups
on Oct 24, 2018 at 4:14:04 pm

Thank you Dan. I know it now.
Have a nice day!


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