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Ps-Design
Known Participant
August 12, 2018
Answered

Group Selected Layers

Is it possible to group all selected layers individually? Each layer would remain within the respective group. All help will be treasured.

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    Stephen Marsh
    Stephen MarshCorrect answer
    Community Expert
    November 3, 2022

    but, but 🙂 can you make it to work for each layer when selected  multiple layers? 


    Try this:

     

    #target photoshop

function main() {

    ///// Process selected layers - from jazz-y /////
    var s2t = stringIDToTypeID;
    (r = new ActionReference()).putProperty(s2t('property'), p = s2t('targetLayersIDs'));
    r.putEnumerated(s2t('document'), s2t('ordinal'), s2t('targetEnum'));
    var lrs = executeActionGet(r).getList(p),
        sel = new ActionReference();
    for (var i = 0; i < lrs.count; i++) {
        sel.putIdentifier(s2t('layer'), p = lrs.getReference(i).getIdentifier(s2t('layerID')));
        (r = new ActionReference()).putIdentifier(s2t('layer'), p);
        (d = new ActionDescriptor()).putReference(s2t("target"), r);
        executeAction(s2t('select'), d, DialogModes.NO);
    //////////

        var layerName = activeDocument.activeLayer.name;
        setFromLayer("[Merge]" + layerName);
        setFromLayer("[Bone]" + layerName);

        ///// Functions /////

        function setFromLayer(layNme) {
            function s2t(s) {
                return app.stringIDToTypeID(s);
            }
            var descriptor = new ActionDescriptor();
            var descriptor2 = new ActionDescriptor();
            var reference = new ActionReference();
            var reference2 = new ActionReference();
            reference.putClass(s2t("layerSection"));
            descriptor.putReference(s2t("null"), reference);
            reference2.putEnumerated(s2t("layer"), s2t("ordinal"), s2t("targetEnum"));
            descriptor.putReference(s2t("from"), reference2);
            descriptor2.putString(s2t("name"), layNme);
            descriptor.putObject(s2t("using"), s2t("layerSection"), descriptor2);
            executeAction(s2t("make"), descriptor, DialogModes.NO);
        }
    }
}
app.activeDocument.suspendHistory("Undo Script", "main()");
    View reply

    21 replies

    Greg von Neulich
    Greg von Neulich
    Known Participant
    May 3, 2023

    we made a Photoshop plug-in that can do this. The unlimited version is 12$. 

     

    Full Version (12$):
    https://exchange.adobe.com/apps/cc/c36e02fb/swift


    Trial Version (Limited to 5 Actions):
    https://exchange.adobe.com/apps/cc/412c1dcd/swift-trial-version

     

    Its also a big help with getting the layers selected.

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    aleksanderk54154416
    Participating Frequently
    November 3, 2022

    Helo!

     

    I need script that will make group from a layer and another group from that group with prefixes.  

    for example result will look like this:

     

    is that possible?

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    Stephen Marsh
    Community Expert
    Stephen MarshCommunity Expert
    Community Expert
    November 3, 2022

    @aleksanderk54154416 – Something like this?

     

    var layerName = activeDocument.activeLayer.name;
setFromLayer("[Merge]" + layerName);
setFromLayer("[Bone]" + layerName);

function setFromLayer(layNme) {
	function s2t(s) {
		return app.stringIDToTypeID(s);
	}
	var descriptor = new ActionDescriptor();
	var descriptor2 = new ActionDescriptor();
	var reference = new ActionReference();
	var reference2 = new ActionReference();
	reference.putClass( s2t( "layerSection" ));
	descriptor.putReference( s2t( "null" ), reference );
	reference2.putEnumerated( s2t( "layer" ), s2t( "ordinal" ), s2t( "targetEnum" ));
	descriptor.putReference( s2t( "from" ), reference2 );
	descriptor2.putString( s2t( "name" ), layNme );
	descriptor.putObject( s2t( "using" ), s2t( "layerSection" ), descriptor2 );
	executeAction( s2t( "make" ), descriptor, DialogModes.NO );
}

     

    A prompt could be added so that you type in the name of the prefix into a GUI, rather than having it hard-coded.

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      aleksanderk54154416
      Participating Frequently
      November 3, 2022

      It works man! I am going to cry now :))

       

      thanx a lot

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      KiriK
      Participant
      September 27, 2019

      Hi all,

       

      Is it possible to make a script or action that both group and name the group the file name at the same time?

      Example:

       

      Layer01

      Layer02

      Layer02

       

      I want to group them and name each group the layer names.

      (exactly the same name as the layer or image)

       

      Group name:

      Layer01

      Layer02

      Layer03

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      Stephen Marsh
      Community Expert
      Stephen MarshCommunity Expert
      Community Expert
      September 27, 2019

      Try this:

       

       

      #target Photoshop

// community.adobe.com/t5/Photoshop/Action-Script-to-group-layers-and-naming/td-p/10635514

// Select all layers - Except Background "Image" Layer
app.runMenuItem(stringIDToTypeID('selectAllLayers')); 

/*
// Select all layers...
var actionDescriptor = new ActionDescriptor();
var actionReference = new ActionReference();
actionReference.putEnumerated( charIDToTypeID( "Lyr " ), charIDToTypeID( "Ordn" ), charIDToTypeID( "Trgt" ) );
actionDescriptor.putReference( charIDToTypeID( "null" ), actionReference );
executeAction( stringIDToTypeID( "selectAllLayers" ), actionDescriptor, DialogModes.NO );
*/

/*
// Select all layers  
// =======================================================  
var idselectAllLayers = stringIDToTypeID( "selectAllLayers" );  
var desc14 = new ActionDescriptor();  
var idnull = charIDToTypeID( "null" );  
var ref8 = new ActionReference();  
var idLyr = charIDToTypeID( "Lyr " );  
var idOrdn = charIDToTypeID( "Ordn" );  
var idTrgt = charIDToTypeID( "Trgt" );  
ref8.putEnumerated( idLyr, idOrdn, idTrgt );  
desc14.putReference( idnull, ref8 );  
executeAction( idselectAllLayers, desc14, DialogModes.NO );  
*/

// community.adobe.com/t5/Photoshop/Group-Selected-Layers/td-p/9983092#10555610
function sTT(v) {return stringIDToTypeID(v)}
(ref = new ActionReference()).putEnumerated
(sTT('document'), sTT('ordinal'), sTT('targetEnum'));
cnt = (lst = executeActionGet(ref).getList(sTT('targetLayersIDs'))).count
for(i = 0; i < cnt;) {
     id = lst.getReference(i++).getIdentifier();
     (ref1 = new ActionReference()).putClass(sTT('layerSection'));
     (dsc1 = new ActionDescriptor()).putReference(sTT('null'), ref1);
     (ref2 = new ActionReference()).putIdentifier(sTT('layer'), id)
     dsc1.putReference(sTT('from'), ref2);
     nme = executeActionGet(ref2).getString(sTT('name'));
     (dsc2 = new ActionDescriptor()).putString(sTT('name'), nme)
     dsc1.putObject(sTT('using'), sTT('layerSection'), dsc2)
     executeAction(sTT('make'), dsc1)
}

       

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        unsandj
        Participant
        September 15, 2021

        Hi, thanks for the script! But could you help me? I want to add a mask to the name. To rename the group like this: [slovo:name_layer]

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        Kukurykus
        Kukurykus
        Legend
        August 12, 2018

        CC 2018:

        function sTT(v) {return stringIDToTypeID(v)}
        (ref = new ActionReference()).putEnumerated
        (sTT('document'), sTT('ordinal'), sTT('targetEnum'));
        cnt = (lst = executeActionGet(ref).getList(sTT('targetLayersIDs'))).count
        for(i = 0; i < cnt;) {
             id = lst.getReference(i++).getIdentifier();
             (ref1 = new ActionReference()).putClass(sTT('layerSection'));
             (dsc1 = new ActionDescriptor()).putReference(sTT('null'), ref1);
             (ref2 = new ActionReference()).putIdentifier(sTT('layer'), id)
             dsc1.putReference(sTT('from'), ref2);
             nme = executeActionGet(ref2).getString(sTT('name'));
             (dsc2 = new ActionDescriptor()).putString(sTT('name'), nme)
             dsc1.putObject(sTT('using'), sTT('layerSection'), dsc2)
             executeAction(sTT('make'), dsc1)
        }
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          Ps-DesignAuthor
          Known Participant
          August 12, 2018

          Phenomenal your script Kukurykus

          You are a genius. Thank you.

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