[Xavier Gomez]

The script Zorro will do most of this i believe.

https://aescripts.com/zorro-the-layer-tagger/

Xavier

[Xavier Gomez]

You can use indexOf, as many times as there are letters to test:

str = thisLayer.textSource.value;
if (str.indexOf("Ó")>=0 || str.indexOf("Ż")>=0 || str.indexOf("Ń")>=0 || str.indexOf("Ć")>=0){
thisValue;
}
else{
thatValue;
};

If you like it short, this also works:

if (RegExp("\[ÓŻŃĆ\]").test(thisLayer.textSource.value)) thisValue else thatValue;

Xavier

[Xavier Gomez]

Not for floating palettes.

[Xavier Gomez]

Which system / AE version do you use to get it to work on a floating palette ?

[Xavier Gomez]

Hi Jacob,

I tried your idea on Windows7 , CC2014, and the event does not trigger.
Only when a child is added, it does trigger, but on the child, not the window itself.

It’s pretty cool idea though, given that .onActivate does not work for dockable panels.
I tried it on a dockable panel instead of a palette, and it seems to work, but again might be version dependant (AE or OS).
This is the code i used:


pal = this instanceof Panel ? this : new Window("palette", "name", undefined, {resizeable:false});

pal.add("button{text: 'button 1'}");
pal.add("button{text: 'button 2'}");
pal.add("button{text: 'button 3'}");

pal.addEventListener("mouseover", mouseoverHandler);
function mouseoverHandler(ev){
if (!ev.view.parent){
this.children[1].enabled = !this.children[1].enabled;
};
};
pal instanceof Window ? pal.show() : pal.layout.layout(true);

(i removed the alert and changed it to something softer, otherwise the event never stops triggering).

Event stuff is ancient Egyptian for me, but this should be roughly true:
Inside the event, ‘this’ is the graphic element the listener was attached to (the window or panel).
ev.view is the graphic element that triggers the event.
So: !ev.view.parent means that the graphic element that triggered the event is a top level container.

Xavier

[Xavier Gomez]

What error do you get ?

[Xavier Gomez]

Sketch is ideal.

I realized some lines in the expression are superfluous and even not used. This one is better:

t1 = 0;
t2 = thisComp.duration;
numCycles = 6;
numStepsPerCycle = 8; // 6+2
// ============================

if (time<=t1 || t2<=time){0;}else{
t = time-t1;
fullDuration = t2-t1;

stepDuration = fullDuration/(numCycles*numStepsPerCycle);
cycleDuration = numStepsPerCycle*stepDuration;
cycleIndex = Math.floor(t/cycleDuration);

phaseShiftPerCycle = 360/numCycles;
shiftStart = cycleIndex*phaseShiftPerCycle;
linear(t%cycleDuration, stepDuration, cycleDuration-stepDuration, shiftStart, shiftStart+360)%360;
};

Xavier

[Xavier Gomez]

Hi,
you can try this, which seems to reproduce your graphics.
But i’m surprised because your graphics has no color doubling as announced in the first post.

t1 = 0;
t2 = thisComp.duration;
numCycles = 6;
numStepsPerCycle = 8; // 6+2
// ============================

if (time<=t1 || t2<=time){0;}else{
t = time-t1;
fullDuration = t2-t1;

stepDuration = fullDuration/(numCycles*numStepsPerCycle);
cycleDuration = numStepsPerCycle*stepDuration;
stepIndex = Math.floor(t/stepDuration);
cycleIndex = Math.floor(stepIndex/numStepsPerCycle);
stepIndexInCycle = stepIndex%numStepsPerCycle;

phaseShiftPerCycle = 360/numCycles;
shiftStart = cycleIndex*phaseShiftPerCycle;
linear(t%cycleDuration, stepDuration, cycleDuration-stepDuration, shiftStart, shiftStart+360)%360;
};

Xavier

[Xavier Gomez]

Thank you for sharing 😉

Can you post a screenshot of how it should look like in the graph editor ?

[Xavier Gomez]

You could also hit P to reveal the position property of the layer, then either :
right click the position property, > reset to recenter,
or shift+scroll the x value to scroll faster.

Xavier