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sensibleworld
sensibleworld
Participating Frequently
July 31, 2017
Answered

Script to Rename Artboards Based on Layers (and Reverse)

  • July 31, 2017
  • 2 replies
  • 3307 views

A couple of years ago I had some awesome help making a couple of scripts to rename layers based on the containing artboard (Re: Script to Rename Artboards with Layer Names). One problem posed (and solved) was:

- Each artboard has a name.

- Each layer NEEDS a name.

- There are an equal number of layers and artboards.

- Only one layer "exists" on each artboard. (i.e. They share coordinates.)

- I would like to name the layer with the same name as its encompassing artboard.

The solution was:

function artboardLayerNameMatch() {  
    if (app.documents.length == 0) {  
        alert("No Open / Active Document Found");  
    } else {  
        var doc, i, l, ab, sel, n;  
        doc = app.activeDocument;  
        for (i = 0, l = doc.artboards.length; i < l; i++) {  
            ab = doc.artboards;  
            doc.artboards.setActiveArtboardIndex(i);  
            doc.selectObjectsOnActiveArtboard();  
            sel = doc.selection[0];  
            sel.parent.name = ab.name;  
            doc.selection = false;  
        }  
    }  
}  
artboardLayerNameMatch();

Now, I'm wondering if the reverse is also possible?

The new scenario is:

  • Each layer has a name.
  • Each artboard NEEDS a name.
  • There are an equal number of layers and artboards.
  • Only one layer "exists" on each artboard. (i.e. They share coordinates.)
  • I would like to name the artboard with the same name as the layer that resides “on” it
This topic has been closed for replies.
Correct answer Disposition_Dev

Here you go. This is certainly not nearly as robust as it could be, but it gets the job done.

function container()
{
    var docRef = app.activeDocument;
    var layers = docRef.layers;
    var aB = docRef.artboards;
    var curLay, curAb, rect, sel, bounds, aBounds, intersect = false;
    var aBLen = aB.length,
        layLen = layers.length;
    function isIntersect(sel, ab)
    {
        return !(sel.l > ab.r || sel.r < ab.l || sel.t < ab.b || sel.b > ab.t);
    }
    if (aBLen === layLen)
    {
        //clear out the selection
        docRef.selection = null;
        //loop the layers
        for (var x = 0; x < layLen; x++)
        {
            curLay = layers;
            curLay.hasSelectedArtwork = true;
            sel = docRef.selection[0];
            bounds = {
                l: sel.left,
                t: sel.top,
                r: sel.left + sel.width,
                b: sel.top - sel.height
            };
            for (var y = 0; y < aBLen && !intersect; y++)
            {
                curAb = aB;
                rect = curAb.artboardRect;
                aBounds = {
                    l: rect[0],
                    t: rect[1],
                    r: rect[2],
                    b: rect[3]
                };
                if (isIntersect(bounds, aBounds))
                {
                    intersect = true;
                    curAb.name = curLay.name;
                }
            }
            intersect = false;
            docRef.selection = null;
        }
    }
    else
    {
        alert("You need to have a matching number of artboards and layers.");
    }
}
container();

2 replies

Disposition_Dev
Disposition_DevCorrect answer
Legend
July 31, 2017

Here you go. This is certainly not nearly as robust as it could be, but it gets the job done.

function container()
{
    var docRef = app.activeDocument;
    var layers = docRef.layers;
    var aB = docRef.artboards;
    var curLay, curAb, rect, sel, bounds, aBounds, intersect = false;
    var aBLen = aB.length,
        layLen = layers.length;
    function isIntersect(sel, ab)
    {
        return !(sel.l > ab.r || sel.r < ab.l || sel.t < ab.b || sel.b > ab.t);
    }
    if (aBLen === layLen)
    {
        //clear out the selection
        docRef.selection = null;
        //loop the layers
        for (var x = 0; x < layLen; x++)
        {
            curLay = layers;
            curLay.hasSelectedArtwork = true;
            sel = docRef.selection[0];
            bounds = {
                l: sel.left,
                t: sel.top,
                r: sel.left + sel.width,
                b: sel.top - sel.height
            };
            for (var y = 0; y < aBLen && !intersect; y++)
            {
                curAb = aB;
                rect = curAb.artboardRect;
                aBounds = {
                    l: rect[0],
                    t: rect[1],
                    r: rect[2],
                    b: rect[3]
                };
                if (isIntersect(bounds, aBounds))
                {
                    intersect = true;
                    curAb.name = curLay.name;
                }
            }
            intersect = false;
            docRef.selection = null;
        }
    }
    else
    {
        alert("You need to have a matching number of artboards and layers.");
    }
}
container();
    sensibleworld
    sensibleworldAuthor
    Participating Frequently
    July 31, 2017

    Works like a charm! Wow. Thank you so much, williamadowling​!

    Disposition_Dev
    Disposition_Dev
    Legend
    July 31, 2017

    Glad to help. Good luck. =)

    Disposition_Dev
    Disposition_Dev
    Legend
    July 31, 2017

    Here's the inelegant, not very efficient pseudo-code. real code to follow:

    if # of layers == # of artboards

         for each layer

              select all artwork on current layer

              bounds = visibleBounds of first item in selection array

              for each artboard

                   if bounds +/- buffer overlaps current artboard

                        current artboard.name = current layer.name

                        break

    else

         alert 'there must be an equal number of artboards and layers'

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