[Dan Ebberts]

Something like this should work:

s = thisComp.layer(“layer with slider”).effect(“Slider Control”)(“Slider”);
linear(s,0,5.2,0,100)

Dan

[Dan Ebberts]

When an expression references the value of the property hosting the expression, it always get the pre-expression value. Expressions have no memory, so an expression can’t access its own results from previous frames. There are usually ways to work around that limitation though, depending on what you’re trying to do.

Dan

[Dan Ebberts]

I guess you could create your own conversion function:


function hexToRgb(theHex){
var r = parseInt(theHex.substr(0,2),16);
var g = parseInt(theHex.substr(2,2),16);
var b = parseInt(theHex.substr(4,2),16);
return [r/255,g/255,b/255,1];
}
c1 = "292B2B";
c2 = "FFFFFF";
v = thisComp.layer("Shape Layer 1").content("Rectangle 1").content("Fill 1").color;
c = ((Math.round(v[0]*255)*256 + Math.round(v[1]*255))*256 + Math.round(v[2]*255)).toString(16);
if (c == c1.toLowerCase())
hexToRgb(c2)
else if (c == c2.toLowerCase())
hexToRgb(c1)
else
value


Dan

[Dan Ebberts]

What version of After Effects are you using? I don’t remember when hexToRgb() came in, but I think it’s fairly recent.

Dan

[Dan Ebberts]

The v variable is the shape’s rgba color array where each color channel is a value between 0 and 1. In the c variable calculation, each color channel value is converted to an integer value between 0 and 255 (by multiplying by 255 and rounding) the red channel is essentially moved 16 bits to the left (by multiplying by 256 twice), the green channel moved 8 bits to the left and the blue channel is left in the right-most 8 bits, then the whole thing is converted to hex so it can be compared to your hex values.

Specifying your colors as rgba arrays would completely change the calculation because you would then be comparing arrays of floating point values rather than hex integers so you might run into issues where the values don’t quite match exactly, so you might have check for values being within a certain range of each other (like .001 or .0001).

Dan

[Dan Ebberts]

Ah. That would be more like this:


c1 = "292B2B";
c2 = "FFFFFF";
v = thisComp.layer("Shape Layer 1").content("Rectangle 1").content("Fill 1").color;
c = ((Math.round(v[0]*255)*256 + Math.round(v[1]*255))*256 + Math.round(v[2]*255)).toString(16);
if (c == c1.toLowerCase())
hexToRgb(c2)
else if (c == c2.toLowerCase())
hexToRgb(c1)
else
value


Dan

[Dan Ebberts]

Like this maybe:


c1 = "292B2B";
c2 = "FFFFFF";
v = value;
c = ((Math.round(v[0]*255)*256 + Math.round(v[1]*255))*256 + Math.round(v[2]*255)).toString(16);
if (c == c1.toLowerCase())
hexToRgb(c2)
else if (c == c2.toLowerCase())
hexToRgb(c1)
else
value


Dan

[Dan Ebberts]

Try it this way:


n = thisLayer("Effects").numProperties;
accum = [0,0];
for (i = 1; i <= n; i++){
if (thisLayer("Effects")(i).name.indexOf("PSR Offset") > -1){
accum += thisLayer("Effects")(i)("Position");
}
}
value + accum


Dan

[Dan Ebberts]

It depends on how you have things set up, but I guess I’d add a dropdown menu control to a layer named “Controls”, add 3 additional elements to the control and name them to match your previous expression:

INSTAGRAM
TWITTER
FACEBOOK
YOUTUBE
SKYPE
SNAPCHAT

Then add an expression like this to each of your logo layers (this example would be for the FACEBOOK layer (#3 in the list):

id = thisComp.layer(“Controls”).effect(“Dropdown Menu Control”)(“Menu”).value;
x = id == 3 ? 630.5 : 572.5; // this is where you specify the id that corresponds to this layer
[x,value[1]]

Something like that…

Dan

[Dan Ebberts]

In my experience, attempting to simulate radio button behavior with checkboxes is doomed to failure. It may seem like it works, but there are subtle order-of-operation dependencies that will make it unreliable. Can’t you use a dropdown menu instead?

Dan