same clipping mask for multiple objects - repeating

Hello ,

It must to be with Script, try this,

***************************

//created by Cmartinez-Gdelgado
//Make sure all artworks must to be in a separate artboards//

var i, ii, doc, count, itemList, countArtboards, artboardIndex;
doc = app.activeDocument;
countArtboards = doc.artboards.length;

for (ii = countArtboards-1; ii >= 0; ii--) {
artboardIndex = doc.artboards.setActiveArtboardIndex(ii);
app.executeMenuCommand("selectallinartboard");
app.executeMenuCommand("makeMask");
}
doc.selection = null;
app.redraw();

*********************************

any cuestion let me know! 🙂

Salut majak!

 Tu règle H à ta mesure et Après sélection des codes barre (image ou groupe)

tu lance le script...

// JavaScript Document for Illustrator
// Réalisation Landry R Wed, 20 May 2020 21:21:52 GMT
// INIT -----
var H = 40;
var dec = 2;
//-----------
var docRef = app.activeDocument;
var selectedItems = selection;
    for (var i = 0; i < selectedItems.length; i++) {
      var item = selectedItems[i];
      var newGroup = layerParent(item).groupItems.add();
          item.move(newGroup,ElementPlacement.PLACEATEND);
      var GBounds = item.geometricBounds;
      var propGobj = prop(GBounds);
      var cadre = newGroup.pathItems.rectangle(GBounds[3]+H,GBounds[0]-dec,propGobj[0]+2*dec, H+dec);
          newGroup.clipped = true;
          newGroup.selected = false;
    }
//--------------------
function layerParent(obj)
{ // détecte le calque parent de obj
  var parent = obj.parent;
  while (parent.typename != "Layer") parent = parent.parent;
  return parent;
}
//---------------
function prop(Bounds) {return [Bounds[2]-Bounds[0],Bounds[1]-Bounds[3]];}
//---------------

de elleere

There may be different approaches to achieve that, but it would be helpful if you were providing at least a couple of sample .ai files, so one may check what could be a reasonably simple way to do what you want to do.

Looks like a candidate for a script.

Are these images? Then maybe a Photoshop action can do the job.