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how del The same name of the channel?

Community Beginner ,
Jan 11, 2017 Jan 11, 2017

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The same name of the channel, is the same pattern. How to keep different channel names, and delete redundant channels with the same name

all channel:1,4,2,1,2,3,1,3,2,4------->channel:1,4,2,3

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Actions and scripting

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correct answers 1 Correct answer

Engaged , Jan 11, 2017 Jan 11, 2017
Hi Roychen, Hope this will help...  var dup=[]; if(app.documents.length>0){ var docRef = app.activeDocument; var mychan = docRef.channels; var l=mychan.length; while(l--) { var mychans=mychan[l].name; if(dup.toString().indexOf (mychans)==-1){ dup.push(mychan[l].name); } else{mychan[l].remove();} } } -yajiv

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Engaged ,
Jan 11, 2017 Jan 11, 2017

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Hi Roychen,

 

Hope this will help... 🙂

var dup=[];
if(app.documents.length>0){
	var docRef = app.activeDocument;
	var mychan = docRef.channels;
	var l=mychan.length;
	while(l--) {
		var mychans=mychan[l].name;            
		if(dup.toString().indexOf (mychans)==-1){
			dup.push(mychan[l].name);
		}
		else{mychan[l].remove();}
	}
}

-yajiv

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Community Beginner ,
Jan 11, 2017 Jan 11, 2017

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hi natrev

very good....thx....

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Community Beginner ,
Jan 12, 2017 Jan 12, 2017

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hi natrev..

have some question,if all channel  is 1,2,3,4,4a,3a,1a,2a,1,2 ,run javscript ,result:1a,2a,3a,4a,1,2...lose 3,4

channel name exmple:

1,1a,1b,1c...

2,2a,2b,2c,2d.2e,2f.....

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Engaged ,
Jan 12, 2017 Jan 12, 2017

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Hi Rochan,

 

Try this code.. 🙂

 

var dup=[];
if(app.documents.length>0){
	var docRef = app.activeDocument;
	var mychan = docRef.channels;
	var l=mychan.length;
	while(l--) {
		var mychans=mychan[l].name;
		if(searchStringInArray (mychans, dup)==-1){ 
			dup.push(mychan[l].name);
		}
		else{mychan[l].remove();}
	}
}

function searchStringInArray (str, strArray) {
	for (var j=0; j<strArray.length; j++) {
		if (strArray[j]==str) return j;
	}
	return -1;
}

 

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Adobe Community Professional ,
Aug 12, 2021 Aug 12, 2021

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LATEST

 

o = {}, chnnls = [].slice.call(activeDocument.channels); while(chnnls.length)
	!o[nme = (shft = chnnls.shift()).name] ? o[nme] = nme : shft.remove()

 

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