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perfectword7
Inspiring
July 11, 2022
Answered

Repeating Expression (not trying to make a looping expression)

  • July 11, 2022
  • 2 replies
  • 436 views

This is a little hard to explain, so if I'm confusing in any way, please let me know so I can clarify.

I have this code expression:

L = thisLayer;
s = thisComp.layer("Controller").effect("Speed")(1)*thisComp.frameDuration;
t2 = key(1).time;
t1 = t2-s;
t4 = key(2).time;
t3 = t4-s;
v1 = thisComp.layer("Controller").effect("Color 1")(1);
v2 = thisComp.layer("Controller").effect("Color 2")(1);
fadeIn = linear(time,t1,t2,v1,v2);
fadeOut = linear(time,t3,t4,v2,v1);

if (time<t1) {v1}
else if (time<t2) {fadeIn;}
else if (time<t3) {v2}
else if (time<t4) {fadeOut;}else{v1}

Currently, this code only works with 2 keyframes. It starts at v1; then Keframe 1 starts a linear transition (length dictated by "s" value) from v1 to v2; then Keyframe 2 starts a linear transition from v2 back to v1.

Here's what I would like to do: I would like to make it such that the code makes it automatically transition from v1 to v2 to v1 to v2 etc. no matter how many keyframes I have (not just with 2 keyframes). Is there a way to do this, or is this simply too crazy to be done?

I appreciate your time. Thanks for any help you can give! 🙂

This topic has been closed for replies.
Correct answer Dan Ebberts

I think it will be something close to this:

s = thisComp.layer("Controller").effect("Speed")(1)*thisComp.frameDuration;
v1 = thisComp.layer("Controller").effect("Color 1")(1);
v2 = thisComp.layer("Controller").effect("Color 2")(1);

n = nearestKey(time).index;
if (time >= key(n).time) n++;
n = Math.min(n,numKeys);
t = key(n).time;
if (n%2){
  linear(time,t-s,t,v1,v2);
}else{
  linear(time,t-s,t,v2,v1);
}

2 replies

Dan Ebberts
Community Expert
Dan EbbertsCommunity ExpertCorrect answer
Community Expert
July 12, 2022

I think it will be something close to this:

s = thisComp.layer("Controller").effect("Speed")(1)*thisComp.frameDuration;
v1 = thisComp.layer("Controller").effect("Color 1")(1);
v2 = thisComp.layer("Controller").effect("Color 2")(1);

n = nearestKey(time).index;
if (time >= key(n).time) n++;
n = Math.min(n,numKeys);
t = key(n).time;
if (n%2){
  linear(time,t-s,t,v1,v2);
}else{
  linear(time,t-s,t,v2,v1);
}
    P
    perfectword7Author
    Inspiring
    July 12, 2022

    This is exactly what I needed! Thank you so much! You're the best.

    Roland Kahlenberg
    Roland Kahlenberg
    Legend
    July 12, 2022

    Look into using Markers to trigger animations/behaviors. They will be a more efficient strategy to achieve what you want. 

    Very Advanced After Effects Training | Adaptive &amp; Responsive Toolkits | Intelligent Design Assets (IDAs) | MoGraph Design System DEV
    P
    perfectword7Author
    Inspiring
    July 12, 2022

    I did consider using markers instead of keyframes; however, I will be using this same code for multiple effects on the same layer, and I want the different effects to transition at different times. If I tie my code to specific markers, wouldn't I have to then have different code for all my different effects? And is there any particular benefit to use markers rather than keyframes to trigger animations/behaviors?

    Also, on another note, I am unable to use the "Responsive Design" features of AfterEffects because I use CS5.5, before that was implemented.

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