Copy link to clipboard
Copied
How to get the angle of an object ļ¼
If you are dealing with most types of art , you can check to see if the automatically-added "BBAccumRotation" tag is at work. This 'tag' is a scripting object that's not accessible from the UI, and likely serves some sort of Illustrator purpose. It is automatically added to art items when the rotation tools are used within AI to rotate them.
#target illustrator
function test(){
var doc = app.activeDocument;
var sel = doc.selection[0];
if(sel.tags.length > 0 && sel.tags[0].name == "BBAccum
...Copy link to clipboard
Copied
For objects which have "matrix" property(images, textframes), the math can be:
var matrix = app.selection[0].matrix;
alert(180/Math.PI * Math.atan2(matrix.mValueC, matrix.mValueD));
Copy link to clipboard
Copied
thanks very much
Copy link to clipboard
Copied
If you are dealing with most types of art , you can check to see if the automatically-added "BBAccumRotation" tag is at work. This 'tag' is a scripting object that's not accessible from the UI, and likely serves some sort of Illustrator purpose. It is automatically added to art items when the rotation tools are used within AI to rotate them.
#target illustrator
function test(){
var doc = app.activeDocument;
var sel = doc.selection[0];
if(sel.tags.length > 0 && sel.tags[0].name == "BBAccumRotation"){
alert(sel.tags[0].value); //in radians
}
};
test();
Copy link to clipboard
Copied
thanks very much
function test(){
var doc = app.activeDocument;
var sel = doc.selection[0];
if(sel.tags.length > 0 && sel.tags[0].name == "BBAccumRotation"){
alert("Angle : "+180 *sel.tags[0].value/Math.PI); //in radians
}
};
test();
Copy link to clipboard
Copied
How to judge a object have been a mirrored?
Copy link to clipboard
Copied
I tried this same code with an object which was reflected over an axis using the Mirror tool (O), and this tag was still added (with no rotation tool used) and the value is negative Pi: -3.141593
Copy link to clipboard
Copied
why?
Copy link to clipboard
Copied
Yes it makes no sense, and I would definitely go the matrix way as moluappleā
Copy link to clipboard
Copied
/////////how to rotate or mirror the image????
ASSET_FOLDER_NAME = "links";
doc = app.activeDocument;
var filename = doc.name.split(".")[0];
asset_folder = new Folder(doc.path + "/" + ASSET_FOLDER_NAME);
if (asset_folder.exists || asset_folder.create()) {
for (i =doc.placedItems.length-1; i >-1; i--) {
linked_file = doc.placedItems.file;
collected_file = new File(asset_folder + "/" + linked_file.name);
if (collected_file.fullName != linked_file.fullName) {
linked_file.copy(collected_file);
doc.placedItems.file = collected_file;
}
}
for (i = doc.rasterItems.length-1; i >-1; i--) {
if (doc.rasterItems.embedded) {
relink_image(doc.rasterItems);
}
}
} else {
alert("Unable to create the assets folder.");
}
function relink_image(myFile){
try{
linked_file = myFile.file ;
tempFpath = new File(asset_folder + "/" + linked_file.name);
if (tempFpath.fullName != linked_file.fullName)
linked_file.copy(tempFpath);
}
catch(err){
var myBBox = myFile.boundingBox;
var myBounds = myFile.geometricBounds;
var tempFpath = newFileName();
var newDocu = app.documents.add();
var newobj = myFile.duplicate(newDocu);
newobj. position = Array(0,0);
if(myBounds[2]-myBounds[0] > myBounds[1]-myBounds[3]){
if(myBBox[2] > myBBox[1]){
newobj.width = myBBox[2]-0.1;
newobj.height = myBBox[1]-0.1;
}else{
newobj.height = myBBox[2]-0.1;
newobj.width = myBBox[1]-0.1;
}
}else{
if(myBBox[2] > myBBox[1]){
newobj.height = myBBox[2]-0.1;
newobj.width = myBBox[1]-0.1;
}else{
newobj.width = myBBox[2]-0.1;
newobj.height = myBBox[1]-0.1;
}
}
var exportPSDOptions = new ExportOptionsPhotoshop();
exportPSDOptions.resolution = 72;
exportPSDOptions.imageColorSpace = ImageColorSpace.CMYK;//gray:1,RGB:2,CMYK:3
newDocu.exportFile (tempFpath , ExportType.PHOTOSHOP,exportPSDOptions);
newDocu.close(SaveOptions.DONOTSAVECHANGES);
}
var replacedImage = doc.placedItems.add();
replacedImage.file = tempFpath;
replacedImage.position = myFile.position;
replacedImage.width = myFile.width;
replacedImage.height = myFile.height;
replacedImage.move(myFile,ElementPlacement.PLACEAFTER);
//how to rotate or mirror the image????
myFile.remove();
}
function newFileName(){
var N = 1;
var loopFlg = true;
while(loopFlg){
var fileObj = new File(asset_folder+"/"+filename+"-"+N+".psd");
N++;
if(! fileObj.exists){
loopFlg = false;
return fileObj;
}
}
}
Copy link to clipboard
Copied
Use the .resize and .rotate to rotate, scale and mirror the object. Using a negative value for the scaleX or scaleY will flip the object over that axis.
Copy link to clipboard
Copied
Hi all,
It seems that in some cases, a rotated item ( a path item in my case ) won't host a BBAccumRotation tag. So all you have left is geometry).
I did write a function to retrieve angle for those shapes but it only work with rectangles as I rely on the 4 points to compute angles. I have absolutely no clue how an angle could be calculated with a non regular shape.
Hoping some brilliant math mind will hang aroundā¦
FWIW, my func for rectangle angle computation:
function getPathItemProperties ( pathitem )
{
var x1, x2, y1, y2, x4,y4, w2,w, h2,h, angleH, cos, factor, angle;
while ( pathitem.typename != "PathItem" )
{
pathitem = pathitem.pageItems[0];
}
x1 = pathitem.pathPoints[0].anchor[0];
y1 = pathitem.pathPoints[0].anchor[1];
x2 = pathitem.pathPoints[1].anchor[0];
y2 = pathitem.pathPoints[1].anchor[1];
x4 = pathitem.pathPoints[3].anchor[0];
y4 = pathitem.pathPoints[3].anchor[1];
w2 = Math.pow ( Math.abs ( x2 - x1 ), 2 ) + Math.pow ( Math.abs ( y2 - y1 ), 2 );
w = Math.sqrt (w2);
h2 = Math.pow ( Math.abs ( x4 - x1 ), 2 ) + Math.pow ( Math.abs ( y4 - y1 ), 2 );
h = Math.sqrt (h2);
angleH = Math.abs ( y2 - y1 );
cos = angleH/w;
factor = ( y2 > y1 )? -1 : 1;
angle = factor * ( Math.acos (cos)*180/Math.PI );
return { w:w*0.3528, h:h*0.3528, r:angle};
}
Copy link to clipboard
Copied
Bonjour Ć tous,
Pour LoĆÆc :
Le fameux tag de nom "BBAccumRotation" est opƩrant sur tous type d'objet.
Sauf que pour certains objets comme une image placƩe par copier coller, un rectangle qui n'a pas encore subit de rotation...
Dans ces cas lĆ , il suffit de crĆ©er le tag par tags.add(), le script qui suit fait ce travail si besoin est.
L'angle est orientĆ©, il est donnĆ© par rapport Ć la direction verticale, dans l'exemple qui suit, l'objet est une instance de symbole, on obtient le mĆŖme rĆ©sultat aprĆØs un "Rompre le lien avec le symbole" (objet groupĆ©s).
Cet angle est mis Ć jour Ć chaque nouvelle rotation (AccumRotation).
exemple :
Grand merci Ć Sily pour avoir abordĆ© ce sujet, car moi aussi je me suis cassĆ© la tĆØte sur ce problĆØme d'orientation sans y parvenir
//
// JavaScript Document for Illustrator
// Finds the tags of name "BBAccumRotation" associated with the selected art item,
// angle to the vertical direction
// of Sily V modif elleere
#target illustratorfunction getAngle(){
var doc = activeDocument;var angle;
if (selection.length == 0) return null;
var sel = doc.selection[0];var nb = sel.tags.length; //alert(nb)
if (!nb) {
if (!confirm ("Create a new Tag with the name BBAccumRotation y/n ?",false,"De Elleere")) return null;
var BBAccumRotationTAG = sel.tags.add();
BBAccumRotationTAG.name = "BBAccumRotation";
BBAccumRotationTAG.value = 0;
nb = 1;
}do {
if (sel.tags[nb-1].name == "BBAccumRotation" ) {
angle = sel.tags[nb-1].value*1; //in radians
angleRad = angle.toFixed(5).replace(/.00000/,"");;
angleDeg = angle*180/Math.PI; //in degrees
angleDeg = angleDeg.toFixed(2).replace(/.00/,"");
alert("mĆ©thode tags\nangleRad = "+angleRad+" rd\nangleDeg = "+angleDeg+"Ā°");
nb = 0
} else{nb--;}
}
while (nb);
return angle;
}getAngle();
Copy link to clipboard
Copied
Hi,
In this snippet, you assume the object isn't rotated explaining the lack of a BBAccumRotation tag. So you add a tag with a 0 rotation angle.
My problem is that when you work with PDF files, even "rotated" object may not have such a tag. See attached screenshots:
Which is wrong obviously.
Unless I missed something?
Copy link to clipboard
Copied
Oui pour un fichier pdf cas particulier.
Copy link to clipboard
Copied
Tag suite:
Maintenant un script qui trace un rectangle autour de l'objet comme le ferait l'outil de sƩlection (V).
exemple 1:
exemple 2:
PS je peux vous faire parvenir un script (par mail) qui sur Illustrator supprime les numƩros de ligne 01. 02.etc pour une rƩcupƩration plus rapide par copier coller.
// JavaScript Document for Illustrator
// Finds the tags of name "BBAccumRotation" associated with the selected art item,
// angle to the vertical direction
// of Sily V modif elleere
#target illustratorfunction getAngle(sel){
var sel, mes, nb, angle, angleRad, angleDeg;
sel = activeDocument.selection[0];
mes = "Create a new Tag with the name BBAccumRotation y/n ?";
nb = sel.tags.length; //alert(nb)
if (!nb) {
if (!confirm (mes,false,"De Elleere")){
return undefined;
}
var BBAccumRotationTAG = sel.tags.add();
BBAccumRotationTAG.name = "BBAccumRotation";
BBAccumRotationTAG.value = 0;
nb = 1;
}do {
if (sel.tags[nb-1].name == "BBAccumRotation" ) {
angle = sel.tags[nb-1].value*1; //in radians
angleRad = angle.toFixed(5).replace(/.00000/,"");;
angleDeg = angle*180/Math.PI; //in degrees
angleDeg = angleDeg.toFixed(2).replace(/.00/,"");
alert("mĆ©thode tags\nangleRad = "+angleRad+" rd\nangleDeg = "+angleDeg+"Ā°");
nb = 0
} else{nb--;}
}
while (nb);
return angle;
}
//---------------
var doc = activeDocument;
if (selection.length) {
var sel, a, pos, Bounds, w, h ,x, y, cadre, newpos, prop0, prop, p0;sel = activeDocument.selection[0];
a = getAngle(sel)*180/Math.PI;
if (!isNaN(a)) {
pos = sel.position;
sel.rotate(-a);
Bounds = sel.geometricBounds;
w = Bounds[2]-Bounds[0];
h = Bounds[1]-Bounds[3];
x = Bounds[0];
y = Bounds[1];
cadre = doc.pathItems.rectangle(y,x,w,h);
cadre.filled = false;
cadre.stroked = true;
cadre.strokeWidth = 1;
sel.rotate(a);
cadre.rotate(a);
newpos = sel.position;
sel.position = pos;
cadre.translate(pos[0]-newpos[0],pos[1]-newpos[1]);
//}
// propriƩtƩs du cadre rectangulaire crƩƩ (De Sily + elleere)
prop0 = {w:w, h:h, r:a};
alert(getArrondi(prop0.r,2)+"Ā°\n"+getArrondi(prop0.w,2)+"\n"+getArrondi(prop0.h,2),"Tag");
p0 = doc.textFrames.pointText(pos);
p0.contents = "Tag\r"+getArrondi(prop0.r,2)+"Ā°\n"+getArrondi(prop0.w,2)+"\n"+getArrondi(prop0.h,2);
p0.translate(0,-h*2);
// propriĆ©tĆ©s du cadre rectangulaire crĆ©Ć© (De LoĆÆc)
prop = getPathItemProperties (cadre);
alert(getArrondi(prop.r,2)+"Ā°\n"+getArrondi(prop.w,2)+"\n"+getArrondi(prop.h,2),"LoĆÆc");
p0 = doc.textFrames.pointText(pos);
p0.contents = "LoĆÆc\r"+getArrondi(prop.r,2)+"Ā°\n"+getArrondi(prop.w,2)+"\n"+getArrondi(prop.h,2);
p0.translate(60,-h*2);
//cadre.remove();}
}
//---------------
function getPathItemProperties ( pathitem )
{
var x1, x2, y1, y2, x4,y4, w2,w, h2,h, angleH, cos, factor, angle;while ( pathitem.typename != "PathItem" )
{
pathitem = pathitem.pageItems[0];
}
x1 = pathitem.pathPoints[0].anchor[0];
y1 = pathitem.pathPoints[0].anchor[1];
x2 = pathitem.pathPoints[1].anchor[0];
y2 = pathitem.pathPoints[1].anchor[1];
x4 = pathitem.pathPoints[3].anchor[0];
y4 = pathitem.pathPoints[3].anchor[1];
w2 = Math.pow ( Math.abs ( x2 - x1 ), 2 ) + Math.pow ( Math.abs ( y2 - y1 ), 2 );
w = Math.sqrt (w2);
h2 = Math.pow ( Math.abs ( x4 - x1 ), 2 ) + Math.pow ( Math.abs ( y4 - y1 ), 2 );
h = Math.sqrt (h2);
angleH = Math.abs ( y2 - y1 );
cos = angleH/w;
factor = ( y2 > y1 )? -1 : 1;
angle = factor * ( Math.acos (cos)*180/Math.PI );
return { w:h, h:w, r:angle} // inversio h et w
}
//---------------
function getArrondi(nb,N)
{ //arrondi nb a N chiffres apres la virgule
return Math.round(Math.pow(10,N)*nb)/Math.pow(10,N);
}
//---------------
Copy link to clipboard
Copied
Oui je te confirme que mĆŖme dernier snippet Ć©choue (mais merci d'avoir proposĆ©).
Ceci dit ela me fait penser que si je rĆ©cupĆØre les points et gĆ©nĆØre un nouvel objet Ć l'identique, peut-ĆŖtre ton snippet fonctionnera.
Copy link to clipboard
Copied
Hello,
AprĆØs une analyse plus profonde, il s'avĆØre que l'objet en question Ć©tait de type "GroupItem" bien qu'il ne soit qu'un tracĆ© unique. En le dĆ©groupant, l'objet PathItem intrinsĆØque dispose bien du tag BBAccumRotation et sa valeur est cohĆ©rente avec l'angle de l'objet.
Donc infine, ton code est probablement trĆØs fonctionnel pour autant que l'on s'adresse au bon objet
Copy link to clipboard
Copied
Hello, Un petite erreur dans le deuxiĆØme script que j'ai corrigĆ©e, (voir plus haut) ligne 13,39,56 accolade dĆ©placĆ©e en 70. On a souvent des surprises avec les pdf non hybrides. J'ai Ć ce propos postĆ© un nouveau sujet sur Mediabox page Illustrator SĆ©lections diffĆ©rentes ? LR