[Dan Ebberts]

Probably something close to this:

tMin = null;
for ( i = 1; i <= thisComp.numLayers; i++){
  if (i == index) continue;
  m = thisComp.layer(i).marker;
  if (m.numKeys == 0) continue;
  n = m.nearestKey(time).index;
  if (time < m.key(n).time) n--;
  if (n > 0){
    t = time - m.key(n).time;
    if (tMin == null){
      tMin = t;
    }else{
      tMin = Math.min(tMin,t);
    }
  }
}
tMin == null ? 0 : tMin

[Dan Ebberts]

Like this:

+[0,20]

[Dan Ebberts]

It’s quite tricky, and it takes a surprising amount of code to do it right. Your best bet might be to get your hands on Jeff Almasol’s rd_Scooter script, which I believe has the code necessary to move keyframes. I think you can still get it from aescripts. The last time I saw it, it was still a .jsx file, not .jsxbin, so you’ll be able to see how he did it.

[Dan Ebberts]

I’m not sure if this is anywhere near what you’re looking for, but this time remapping expression should hold roughly 20% of the frames for 3 frames, the others for 2:

randomPct  = 20;
f = timeToFrames(time);
fNew = i = 0;
seedRandom(index,true);
framesToTime(fNew)
while (i <= f){
  i += random(100) < randomPct ? 3 : 2;
  fNew ++;
}
framesToTime(fNew)

[Dan Ebberts]

This example shows how you could add [10,10] to each vertex of a shape path:

 var myComp = app.project.activeItem;
	var myLayer = myComp.layer(1);
	var myPathObject = myLayer.property("Contents").property("Shape 1").property("Contents").property("Path 1").property("Path");
	var myPath = myPathObject.value;
	var myVertices = myPath.vertices;
	var v;
	for (var i = 0; i < myVertices.length; i++){
		v = myVertices[i];
		myVertices[i] = [v[0] + 10, v[1] + 10];
	}
	myPath.vertices = myVertices;
	myPathObject.setValue(myPath);

[Dan Ebberts]

Are you talking about scripting? Unfortunately, you can’t change the time of a keyframe via scripting. You have to create a new keyframe, copy all the attributes of the old one (which can be quite involved and order-sensitive), and delete the old one.

[Dan Ebberts]

Hard to say for sure, without seeing what you have, but I’m guessing that this:

if(ManuLang == 5 ){100} else {0};
if(dropMenu == 4 ){100} else {0};

needs to be like this:

if((ManuLang == 5 ) && (dropMenu == 4)){100} else {0};

[Dan Ebberts]

I think what you’re trying to do falls in the category of speed control described here:

http://www.motionscript.com/articles/speed-control.html#linear

[Dan Ebberts]

Something like this for Comp B should get the text from Comp A, but replace the line breaks with spaces:

txt = comp("Comp A").layer("Master Text").text.sourceText;
txt.split("\r").join(" ")

[Dan Ebberts]

This is pretty limited and not well tested. If the two paths don’t have the same number of points, it doesn’t do anything. It assumes that the slider is on the layer with the expression.

path1 = thisComp.layer("Shape Layer 1").content("Shape 1").content("Path 1").path;
path2 = thisComp.layer("Shape Layer 2").content("Shape 1").content("Path 1").path;
s = effect("Slider Control")("Slider");
p1 = path1.points();
p2 = path2.points();
it1 = path1.inTangents();
it2 = path2.inTangents();
ot1 = path1.outTangents();
ot2 = path2.outTangents();
if (p1.length == p2.length){
  newP = [];
  newIt = [];
  newOt = [];
  for (i = 0; i < p1.length; i++){
    newP.push(linear(s,0,100,p1[i],p2[i]));
    if (it1.length > 0 && it2.length > 0){
      newIt.push(linear(s,0,100,it1[i],it2[i]));
    }else if (it1.length > 0){
      newIt.push(linear(s,0,100,it1[i],[0,0]));
    }else if (it2.length > 0){
      newIt.push(linear(s,0,100,[0,0],it2[i]));
    }
    if (ot1.length > 0 && ot2.length > 0){
      newOt.push(linear(s,0,100,ot1[i],ot2[i]));
    }else if (ot1.length > 0){
      newOt.push(linear(s,0,100,ot1[i],[0,0]));
    }else if (ot2.length > 0){
      newOt.push(linear(s,0,100,[0,0],ot2[i]));
    }
  }
  createPath(newP,newIt,newOt,path1.isClosed());
}else
  value;