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get scale (h, v)

Community Beginner ,
Feb 20, 2018 Feb 20, 2018

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Hi,

I need help, I need to know the scale percentage of the image placed in illustrator. I can get it from matrix,  but after I rotate the image,  everything changed. What I need is exactly the same what I see in Link Information  " scale (h, v)"    - this when you double click the image from the link panel.

Thank you.

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Valorous Hero ,
Feb 21, 2018 Feb 21, 2018

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Seems like nothing works best than matrices:

alert("Horizontal Scale : "+Math.round((app.selection[0].matrix.mValueA*100)*100)/100+"\r"+"Vertical Scale : "+Math.round((app.selection[0].matrix.mValueD*100)*100)/100+"%");

Capture d’écran 2018-02-21 à 16.29.26.png

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Community Beginner ,
Feb 22, 2018 Feb 22, 2018

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Thanks for the help!!!

But if forgot to mention that I'm using applescript .

Anyway, I found this...    do shell script "ruby -e 'puts Math.atan2(" & mvalue_b of m & "," & mvalue_a of m & ") / Math::PI * 180'"  by this will give the rotation percentage of the image.

So I use it to rotate back my image to 0 deg to get the right scale percentage:

set z to selection

set x to do shell script "ruby -e 'puts Math.atan2(" & mvalue_b of m & "," & mvalue_a of m & ") / Math::PI * 180'"

set y to get mvalue_b of matrix of the selection

                      

if y is not equal to 0 then

rotate z angle x about top right

set z to get mvalue_a of matrix of the selection

undo

else

set z to get mvalue_a of matrix of the selection

end if

this works for me, but if you can still give me a better syntax, ill appreciate.  Thank you!

Jhon

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Adobe Community Professional ,
Feb 22, 2018 Feb 22, 2018

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Does applescript not have the trig math operations?

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Adobe Community Professional ,
Feb 22, 2018 Feb 22, 2018

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Hi Loic,

I agree to your solution. However, we can't get correct scale when it had a rotatation angle like below.

スクリーンショット 2018-02-23 11.14.33.png

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Adobe Community Professional ,
Feb 22, 2018 Feb 22, 2018

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Here is a PoC code, get scale from rotated image.

var tg = app.selection[0];

var mx = tg.matrix;

var deg = Math.atan2(mx.mValueB, mx.mValueA) * 180 / Math.PI;

var nm = new Matrix;

nm.mValueA = nm.mValueD = 1;

nm.mValueB = nm.mValueC = 0;

var rtmx = app.concatenateRotationMatrix(nm, deg);

var sc = mx.mValueA / rtmx.mValueA;

alert(sc);

I'm stacking math a few month ago. Thank you for your info jhonperezph!

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Enthusiast ,
Feb 26, 2018 Feb 26, 2018

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Bonjour,

merci Ten A

// JavaScript Document Illustrator

// elleere Sun, 25 February 2018 21:08:12 GMT

{

  if (selection) {

  var sel = selection[0];

    if(sel.typename=='RasterItem' || sel.typename == 'PlacedItem') {

      var propT =  getscale1(sel);

      alert("scale1(H,V) : ("+propT.H+"%, "+propT.V+"%), Rotation : "+propT.A+" °");

      var propT =  getscale2(sel);

      alert("scale2(H,V) : ("+propT.H+"%, "+propT.V+"%), Rotation : "+propT.A+" °");

    }

  }

//---------------

  function  getscale1(tg) {

    var mtx, deg, degArd, scH, scV;

        mtx = tg.matrix;

        deg = Math.atan2(mtx.mValueB, mtx.mValueA) * 180 / Math.PI;

        degArd =getArrondi(deg,3);

        tg.rotate(-deg);

        mtx = tg.matrix;

        scH = getArrondi(mtx.mValueA * 100,3);

        scV = getArrondi(mtx.mValueD * 100,3);

        tg.rotate(deg);

        return {H:scH,V:scV,A:degArd}

    }

  function getscale2(tg){

    var mx, deg, degArd, nm, rtmx,scH, scV;

        mx = tg.matrix;

        deg = Math.atan2(mx.mValueB, mx.mValueA) * 180 / Math.PI;

        degArd =getArrondi(deg,3);

        nm = new Matrix;

        nm.mValueA = nm.mValueD = 1;

        nm.mValueB = nm.mValueC = 0;

        rtmx = app.concatenateRotationMatrix(nm, deg);

        scH = getArrondi(mx.mValueA / rtmx.mValueA * 100,3);

        scV = getArrondi(mx.mValueD / rtmx.mValueD * 100,3);

        return {H:scH,V:scV,A:degArd}

  }

  //---------------

  function getArrondi(nb,N) {

  //arrondi nb a N chiffres apres la virgule

  return Math.round(Math.pow(10,N)*nb)/Math.pow(10,N);

  }

//---------------

}

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Adobe Community Professional ,
Feb 26, 2018 Feb 26, 2018

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Bon travail!

Merci de l'avoir partagé.

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Enthusiast ,
Feb 26, 2018 Feb 26, 2018

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LATEST

Rectificatif pour le script donné plus haut :

Inverser la valeur de l'angle

deg = -Math.atan2(mtx.mValueB, mtx.mValueA) * 180 / Math.PI;

scV valeur absolue

scV = Math.abs(getArrondi(mx.mValueD / rtmx.mValueD*100,3));

Désolé LR

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