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pull layer’s comp name?
Hey all, quick expression question:
If I have declared a layer in a variable:
L = comp("Comp 1").layer("Shape Layer 1");
can I later pull the comp name for L? e.g. (doesn’t work)
N = L.propertyGroup(1).nameor do I need to declare them individually?
C = comp("Comp 1");
L = C.layer("Shape Layer 1");
n = C.name;
I would really prefer to declare as a single line (as this will half the time be pickwhip’d, half not), so if it’s at all possible I’d love to hear!Alex Printz
Mograph Designer -
4 Replies
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The second one. Expressions don’t have anything like the containingComp attribute that’s available in scripting.
Dan
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Actually, it turns out that you can. After a bit of experimenting, you can grab the comp :
L = comp("Comp 1").layer("Shape Layer 1");
n = L.thisComp.name;Strangely, if you do this, you cannot grab the layer the same way:
R = comp("Comp 1").layer("Shape Layer 1").content("Rectangle Path 1");
n = R.thisLayer.namebut you can do this:
R = comp("Comp 1").layer("Shape Layer 1").content("Rectangle Path 1");
n = R.propertyGroup(2).nameAlex Printz
Mograph Designer -
>n = L.thisComp.name;
I’m really surprised that this works. The expression language reference lists thisComp as a global object representing the comp containing the expression, but it must really be a layer attribute, with the layer defaulting to thisLayer.
thisLayer.thisComp.name
(which I would expect to throw an error) gives the same result as:
thisComp.name
You learn something every day I guess…
Dan
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Yeah that was really unexpected now that I’m looking at adobe reference page. Interesting that it can be referenced via the layer. Might have to play around with that some more.
Alex Printz
Mograph Designer
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