tiscavalcanti

I'm not sure what your question is. Did you post a conversation between you and someone else or something?

I myself will continue trying to do. That is it.

var x = 0;var y = 0;var b = 2;var c = 3;So;first function of x = valueXOffset (xFunctionFirst) // OffsetIf b = 1return{96 // 5% of 1920}y = 54 // 5% of 1080Else if b = 2return{192 // 10% of 1920}y = 108 // 10% of 1080Else if b = 3return{288 // 15% of 1920}y = 162 // 15% of 1080Else b = (any value)return{0}y = 0second function of x = valueXFinal (xFunctionSecond)If c = 1return{xFunctionFirst}Else If c = 2return{1920/2}Else If c = 3return{1920 - xFunctionFirst}Else c = (any value)0call function here, to get x value and y too.[x, y];

The restrictions I have, it is to be done with functions, please.

 

You can have your function return an array, but unless you're operating a and z together in two different ways, I don't see a reason to do it.

 

Returning both values with an array:

function val(a, z) {  return [ a/2, z+8 ];}

A better approach:

function half(x) {  return x/2;}function plus8(x) {  return x + 8;}

Thank you!

I have:

function val(a,z){

return a/2, z+8;

}

val(3,4);

Put this in a Array;

x = 1,5

y = 12

[x,y];

Thanks.

w = thisComp.width;h = thisComp.height;s = 0;p = [.95, .90, .85, .15, .10, .05];x = value[0];y = value[1];function getLayerName(theLayer,theName,theParm){  try{    return theLayer.effect(theName)(theParm).name;  }catch(err){    return "None";  }}L = thisComp.layer("Less");a = getLayerName(L,"Scale","Layer");b = getLayerName(L,"Position","Layer");c = getLayerName(L,"Side","Layer");

This is it.

It still isn't clear to me what code you are replacing with this loop.

 

You say...

 

 

...so show us a dozen lines of code.

Thank you too! I'm able.

You can set up an object like this and then use it to populate the variables with the values:

var obj = {  a: 'Scale',  b: 'Position',  c: 'Side'};

You could loop through the properties of that object and use them to set the variables with their corresponding values, but doing that is not going to improve efficiency. It might make the code shorter, but it's more efficient to just execute each statement individually rather than looping.

 

Thank you! I understand.

It still isn't clear to me what code you are replacing with this loop.

 

You say...

 

 

...so show us a dozen lines of code.

Oh, yeah!

I want to optimize this expression.

// Position    // Variables:  var w=1920;     // Comp Width  var h=1080;     // Comp Height  var p1=.95;     // variable to multiply by 95%  var p2=.90;     // variable to multiply by 90%  var p3=.85;     // variable to multiply by 85%  var p4=.15;     // variable to multiply by 15%  var p5=.10;     // variable to multiply by 10%  var p6=.05;     // variable to multiply by 5%  var s;          // variable to be modified  var x;          // variable to represent Width Property value  var y;          // variable to represent Height Property value    // Variables with Try/Catch Block to Prevent errors:    // (I wanted to simplify the part below, making it more optimized.)    try{  a=thisComp.layer("Less").effect("Scale")("Layer").name;     // Variable = "Scale" - Layer Control Effect  }catch(err){  a=0;        // Prevention for when the effect is selected "None".  };  try{  b=thisComp.layer("Less").effect("Position")("Layer").name;      // Variable = "Position" - Layer Control Effect  }catch(err){  b=0;        // Prevention for when the effect is selected "None".  };  try{  c=thisComp.layer("Less").effect("Side")("Layer").name;      // Variable = "Side" - Layer Control Effect  }catch(err){  c=0;        // Prevention for when the effect is selected "None".  };    // Conditions:    // (I wanted to simplify these conditions with functions or loops, for example.)    if(thisComp.layer("Less").effect("Master Control")("Checkbox")==0){     // Condition to when Checkbox is disabled, the property back to original value  value;  }else{      // And otherwise  if(b=="Less"){      // If "Position" - Layer Control Effect = Layer "Less"  y=h*p1;     // Height value = Comp height * 95% | Result = 1026  s=w*p6;     // variable s = Comp width * 5% | Result = 96  }else if((b=="Half") || (b=="Default")){        // Otherwise, If "Position" - Layer Control Effect = Layer "Half" or "Default"  y=h*p2;     // Height value = Comp height * 90% | Result = 972  s=w*p5;     // variable s = Comp width * 10% | Result = 192  }else if(b=="More"){        // Otherwise, If "Position" - Layer Control Effect = Layer "More"  y=h*p3;     // Height value  = Comp height * 85% | Result = 918  s=w*p4;     // variable s = Comp width * 15% | Result = 288  }else{      // And otherwise  y=value[1];     // Height value = Original value  };  if(a==0){       //  If "Scale" - Layer Control Effect = None  x=w/2;      // width value = Comp Width / 2 | Result = 960  }else{      // And otherwise  if((c=="Less") || (c=="Default")){      // If "Side" - Layer Control Effect = Layer "Less" or "Default"  x=s;        // Width value = variable s  }else if(c=="Half"){          // Otherwise, If "Side" - Layer Control Effect = Layer "Half"  x=w/2;      // Width value = Comp width / 2 | Result = 960  }else if(c=="More"){          // Otherwise, If "Side" - Layer Control Effect = Layer "More"  x=w-s;      // Width value = Comp width - variable s  }else{      // And otherwise  x=value[0];     // Width value = Original value  };  };  [x,y]       // Property | Array = Final Value of Width and Height  };  

That looks like code from some sort of animation framework. What exactly is it? Adobe After Effects?

Is there a way to do this more easily?

How to add a value in the middle of a code line?

And there is a way to separate each line of output of a for loop?

Thanks.

That looks like code from some sort of animation framework. What exactly is it? Adobe After Effects?

What I want is to actually simplify the code, to avoid repetition.

That looks like code from some sort of animation framework. What exactly is it? Adobe After Effects?

thisComp.effect(Scale)('Layer').namethisComp.effect(Position)('Layer').namethisComp.effect(Side)('Layer').name

That looks like code from some sort of animation framework. What exactly is it? Adobe After Effects?

Yeah, Adobe After Effects.

Can you show us a small example of this code in a non-looping form that actually works?

var a = thisComp.effect("Scale")("Layer").name;  var b = thisComp.effect("Position")("Layer").name;  var c = thisComp.effect("Side")("Layer").name;  

I don't really understand what you're saying, but you can use the +, -, *, and / operators to do basic math.

Good, you is right, Thank You!

That MDN site has examples for all of the methods in the Math object. The other one is just some subtraction and division. Get started and ask questions if you get stuck.

I got. Examples:

.20*76 | 20% of 76 = 15.2

1.20*76 | 76 + 20% = 91.2

(1-.20)*76 | 76 - 20% = 60.8

Calculate numerical difference between two values:

(-8)+(3) = - 5 | balance value.

This is it. Thanks.

That MDN site has examples for all of the methods in the Math object. The other one is just some subtraction and division. Get started and ask questions if you get stuck.

Okay.

If you're trying to calculate the "distance" between two numbers as a positive value, that's the absolute value. The Math object has a method for it.https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/MathTo get the percentage difference you would take the absolute value of the difference and then divide the original number by that.

You can take some examples, please?

Thank you!

I need make a calculation which allow know difference between two numbers. I can subtract the smallest of the larger, but this calculation is very limited, implies if a number is negative and the other positive.

I want to know another way to do this.

• It is possible to do this calculation in JavaScript through the:

1. modulo;
2. function;
3. (Others);

Mainly:

Another important thing which I want, is to calculate percentage difference through two differents numbers. To know the value resultant.

• I want to know all the possible ways to do this. Because I came across in an embarrassing situation in a video Software that uses expressions based on the JavaScript language.

Thanks!