[Dan Ebberts]

If you’re animating the Trim Path’s End property with linear keyframes, there’s likely a more efficient way to do it.

[Dan Ebberts]

The key to something like this is for the expression to calculate when the Trim Paths reached the triggering threshold for this particular layer and calculate the time since that event, and use that to drive the animation. Worst case, you could do something like this:

dur = .5; // duration of scale-up (in seconds)
s = 0;
x = thisComp.layer("Timeline Animation").effect("Trim Path to Pixels")("Slider");
if (x > transform.xPosition){
  t = time - thisComp.frameDuration;
  while (x.valueAtTime(t) > transform.xPosition){
    t -= thisComp.frameDuration;
  }
  delta = time - t;
  s = linear(delta,0,dur,0,100);
}
[s,s]

[Dan Ebberts]

I’m not sure, but you might be after something like this:

effect(effect("Nslider").propertyIndex-1).name

[Dan Ebberts]

It’s been a while since I’ve done much with plugin code, but I seem to recall that when you add a new control to your plugin, you just need to make and assign it a new, unique id and it should be backwards compatible. It sounds like you may have had a different experience.

[Dan Ebberts]

I think the problem might be that when xfreq and yfreq point to individual x and y values of the point control, xfreq.numKeys and xfreq.key() no longer make sense.

[Dan Ebberts]

You’ll need a recursive property sniffer to check for expression errors. Something like this should be close:

var foundOne;
function processProperty(theProp){
	if (theProp.propertyType == PropertyType.PROPERTY){
		if (theProp.canSetExpression && theProp.expressionError != ""){
			foundOne = true;
		}
	}else{ // must be a group
		for (var i = 1; i <= theProp.numProperties; i++){
			if (! foundOne) processProperty(theProp.property(i));
		}
	}
}
var myComp, myLayer;
for (var i = 1; i <= app.project.numItems; i++){
	if (app.project.item(i) instanceof CompItem){
		myComp = app.project.item(i);
		for (var j = 1; j <= myComp.numLayers; j++){
			myLayer = myComp.layer(j);
			if (! myLayer.locked){
				foundOne = false;
				processProperty(myLayer);
				if (foundOne){
					myLayer.property("Effects").addProperty("ADBE Fill");
				}
			}
		}
	}
}

[Dan Ebberts]

I think you’d replace the last line with this:

s = Math.max(xclamp,yclamp);
[s,s]

[Dan Ebberts]

The only other (besides sourceRectAtTime()) tool I know of is sampleImage(). It’s really inefficient because it has to do a lot of calculations, but you can use it to find the non-zero alpha extents of your layer. This expression will give you the left, right, top, and bottom extents. You’ll need to tailor it to your exact requirements.

L = thisComp.layer("target");
h = L.height/2;
w = L.width/2;
for (i = 0; i < L.width; i++){
  if (L.sampleImage([i,h],[.5,h])[3]>0) break;
}
left = i;
for (i = L.width-1; i >= 0; i--){
  if (L.sampleImage([i,h],[.5,h])[3]>0) break;
}
right = i;
for (i = 0; i < L.height; i++){
  if (L.sampleImage([w,i],[w,.5])[3]>0) break;
}
top = i;
for (i = L.height-1; i >= 0; i--){
  if (L.sampleImage([w,i],[w,.5])[3]>0) break;
}
bottom = i;

[Dan Ebberts]

This should work:

time*Math.pow(10,effect("slider 2")("Slider"))

[Dan Ebberts]

Hmmmm. I can’t picture how that would work. So the expression for each control would include its own keyframed value plus the appropriate symmetrical controls? I think that would set up the kind of viscous cycle loop that expressions can’t resolve. Kind of like when you try to create a radio button control out of checkboxes. The expressions end up fighting each other.