[Filip Vandueren]

It appears it has left and right paragraph margins set up, or the text box was not created to be exactly 1000px wide maybe?

[Filip Vandueren]

See here:

https://creativecow.net/forums/thread/paragraph-bouding-control/

Paragraph bouding control

[Filip Vandueren]

You are correct that for Z, you could still use the actual rotation as before.

To correct the 2D position of the null I would have to see the entire project, because it depends on the positions of face and null layer and wether the heads position is animated etc.

[Filip Vandueren]

Here’s how I would do it:

https://imgur.com/7q9K73k

View post on imgur.com

[Filip Vandueren]

Hi Fabrice, this was an accidental double post of this thread:

https://creativecow.net/forums/thread/text-highlight-expression-selector/

Text highlight (expression selector)

[Filip Vandueren]

Maybe all you need is a lookAt() expression on the orientation of the layer?

[Filip Vandueren]

You should send a 2D or 3D value into toWorld(), usually that doesn’t need to be the position of the layer, but a point on the layer. Most often we want to find the position of a layer’s anchorPoint in space, and for a default null that would be [0,0,0].

So the code for what you’re asking is this:

nullLayer = comp("comp2").layer("2D_null");
xyz=thisLayer.fromWorld(nullLayer.toWorld([0,0,0]));
x=xyz[0];
y=xyz[1];

But I’m not entirely sure this is going to do what you expect it to do. When do you expect the rotations to be at 0 ?

[Filip Vandueren]

If you want 1 match per name max, then change the regexp from “gi” to just “i”

[Filip Vandueren]

I think this would do it:

names = thisComp.layer("A").text.sourceText.replace("\u200B","").split(","); // this makes it an Array, (after removing any invisible spaces)
word = thisComp.layer("B").text.sourceText.value;
regex = new RegExp(word, "gi"); // ,"i" makes it a case insensitive search
if (word.length > 1) {
	names
		.filter(a => a.match(regex)) // only array elements that match the word
		.map(a => a.trim().replace(regex, "\u200B$&\u200B")) // trim extra whitespace, then surround the found queries with invisible spaces
		.sort( (a,b) => a.toLowerCase().indexOf(word.toLowerCase()) - b.toLowerCase().indexOf(word.toLowerCase()))
		.join("\r"); // turns the array into a string 
} else {
	"";
}

It sorts by where in the name the substring is found, and for equal positions, should retain the alphabetical sorting that was already in place.

all the toLowerCase() makes it a bit messy, but that’s to make that part of the code case insensitive too. Could be done with a regular expression too perhaps.

[Filip Vandueren]

Yes, the expression seems to have been mangled by posting it here.

The for loop should be this:

for (i=1; i<textindex; i++) {

if (txt[i-1]=="\u200B") sel = !sel;
}

it’s certainly possible to do an extra custom sort() on the array. Let me think about how to phrase that in a function.