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pixxxelschubser

Community Expert

pixxxel schubser

active since 2009 | 6100+ posts

pixxxelschubser

Community Expert

pixxxel schubser

active since 2009 | 6100+ posts

Activity

Hmm, this is the scripting forum, not the main forum. wobblewoo wrote: … Im guessing i save those actions into a .js file and then call upon that script somehow … into a .jsx file in the right scriptfolder (for your PS-version) close and reopen PS open a (copy of a) file and menu --> file --> scripts (or similar, I only have a german CS3-Version)

wobblewoo wrote: … but then name the layer the filename and make it a smart object … For this part try this: // docNameAsActiveLayerNameAndMakeSO.jsx var aLay = app.activeDocument.activeLayer aLay.name = app.activeDocument.name; var idSO = stringIDToTypeID( "newPlacedLayer" ); executeAction( idSO, undefined, DialogModes.NO ); Have fun

It's tricky. // GradientBlackWhite_rotated.jsx // requires an open document and one selected object var newGradient = app.activeDocument.gradients.add(); var colorOfGradient = new GradientColor(); var myGradientAngle = 80; colorOfGradient.gradient = newGradient; doc.selection[0].fillColor = colorOfGradient; /*Thanx to John Wundes for this -->*/doc.selection[0].rotate(myGradientAngle, false, false, true, false, Transformation.CENTER); redraw(); alert("gedreht"); Have fun

Sorry, this is like Dinner for One: „same procedure as last year“ http://forums.adobe.com/message/4380742#4380742

Oh no, what have I done? That should be a joke. As response to the contribution of Michael L Hale. Michael L Hale wrote: … For example myColor = new SolidColor(); backgroundColor = myColor; will throw the same internal error message … He has right. A solid color hex value is default 'FFFFFF' But you can pass the default value to the background color and fill the document with the correct syntax like this: myColor = new SolidColor(); backgroundColor.rgb.hexValue = myColor.rgb.hexValue; app.activeDocument.selection.fill( backgroundColor, ColorBlendMode.NORMAL, 100, false );

Hello Michael L Hale, Michael L Hale wrote: … But it is really not that useful to use the defalut because it doesn't work with all Photoshop methods or properties. For example myColor = new SolidColor(); backgroundColor = myColor; will throw the same internal error message … understand fun? //MakeNewDocFillWithBGColor.jsx var Syntax = false; var docRef = app.documents.add(300, 300, 72, null, NewDocumentMode.RGB, DocumentFill.TRANSPARENT); var selRef = app.activeDocument.selection; var myAlert = null; try { myColor = new SolidColor(); backgroundColor = myColor; Syntax = true; selRef.fill( backgroundColor, ColorBlendMode.NORMAL, 100, false ); myAlert = "No No No"; } catch (e) { $.sleep (999); if (Syntax == false) { var MakeSyntax = true; myColor = new SolidColor(); backgroundColor.rgb.hexValue = myColor.rgb.hexValue; selRef.fill( backgroundColor, ColorBlendMode.NORMAL, 100, false ); myAlert = "Script runs, if MakeSyntax == "+MakeSyntax; } } alert(myAlert,"Have Fun") greetings

You're welcome. Additional you can the foreground color use to paint or fill or stroke selections. var myColor = app.foregroundColor;

Hello Radsvid789, your script: var myColor = new SolidColor(); app.documents.add( 100, 100, 72, "tmp", NewDocumentMode.RGB, DocumentFill.TRANSPARENT ); app.activeDocument.selection.selectAll(); app.activeDocument.selection.fill( myColor ); cannot run. At first you must define the new SolidColor e.g. like this: var myColor = new SolidColor(); myColor.rgb.red = myColor.rgb.green = myColor.rgb.blue = 0; Try this one: var myColor = new SolidColor(); myColor.rgb.red = myColor.rgb.green = myColor.rgb.blue = 0; app.documents.add( 100, 100, 72, "tmp", NewDocumentMode.RGB, DocumentFill.TRANSPARENT ); app.activeDocument.selection.selectAll(); app.activeDocument.selection.fill( myColor ); The same problem in your second example. Have fun

Hello [Jongware] I'm slightly confused. e.g. 4500 pixels x 4500 pixels are 20,250,000 pixels. Thats IMHO more than 20 MP, or not?

That's right. Only a spot color can be LAB. The same in the UI.

You have an opened document with page items? try this: app.activeDocument.layers[0].pageItems[0].name = "my Item #1"; app.activeDocument.layers[0].pageItems[0].selected = true; app.redraw(); alert(app.activeDocument.layers[0].pageItems[0].name);

Hello booyajamo, another option is try & catch Here is an example: var Num = 1; // Take a smaller integer, but greater than zero var stuff = prompt ("Your Choice (Only a real number > 0 will run without an exit)", Num); try { for (i=0; i<stuff; i++){ app.activeDocument.layers[0].textFrames.add();} app.activeDocument.layers[0].textFrames[stuff].contents = "Your choice was a good choice ;)"; // when stuff isn't a number --> then exit here // … all your good stuff alert ("Finish - Your choice was a good choice ;)"); } catch (e) { alert ("No good choice - sorry for exit"); } Test it. Good luck

if ( app.documents.length > 0 ) { var numberOfItems = 3; var selectedItem = app.activeDocument.selection; for (i=0; i < numberOfItems; i++) { newItem = selectedItem.duplicate(); } } maguskrool wrote: Thanks. I got an error using this option, though, not sure why but it says the last item (in this case, selectedItem when i = 2) is undefined. Your script says: When (min) one document is open --> say numberOfItems = 3 //the counter is 3 --> min three or more items should be selected by yourself selectedItem --> is the array of all selected items in document then go for counter = 0 to counter < numberOfItems (means 3-1) --> counter 0 = loop 1 --> duplicate this (item 1 of 3 selected) item counter 1 = loop 2 --> duplicate this (item 2 of 3 selected) item //when only 1 item is selected by yourself --> result is undefined counter 2 = loop 3 --> duplicate this (item 3 of 3 selected) item //when only 2 items are selected by yourself --> result is undefined It's equal, how many items are selected by user - but min you have defined here --> var numberOfItems = 3; You need in this example a selection of three items (better: three selected items)

the other way (your script): if ( app.documents.length > 0 ) { var numberOfItems = 3; var selectedItem = app.activeDocument.selection; for (i=0; i < numberOfItems; i++) { newItem = selectedItem.duplicate(); } }

@ mauro8282, why is your question (posting # 2) at the same time the correct answer?

Hello, I prefer the slice-method. Take a look at this example: // at first - open a document, then run script var aDoc = app.activeDocument; var aName = aDoc.layers[0].name; alert(aName); var newName = "LayerXYZ"; // Changing the new name of top level aDoc.layers[0].name = newName; // display the new name | the result will be: LayerXYZ alert ("The new name of this Layer is: " +newName); // shorter Variable newName = aDoc.layers[0].name.slice(0,-3); // Changing the new name of Layer 1 aDoc.layers[0].name = newName; // display the new name | the result will be: Layer alert("The new short name of this Layer is: " +newName);

Please, show it to us

Hi CarlosCanto, this is funny. The same answer I gave to the thread opener at noon today in another forum: http://www.hilfdirselbst.ch/foren/Print_Probleme_im_Illustrator_-_Scripte_P506095.html?sb=post_latest_reply;so=ASC;forum_view=forum_view_collapsed;;page=unread#unread

Loop through subfolders e.g. like this: http://forums.adobe.com/message/3139217#3139217

That would be the easiest option: #target Illustrator // Remove9of10.jsx // graph must be breaked; // only if category axis is the x-axis; // all selected tick marks must be ungrouped; // http://forums.adobe.com/thread/1095789?tstart=0 // the first tick mark (normaly on position 0 in graph) is also deselected/removed // by pixxxelschubser var myDoc = app.activeDocument; var Sel = myDoc.selection; var SelLen = Sel.length; var p1 = Sel[0].position[0]; var p2 = Sel[SelLen-1].position[0]; var x = 10; x = prompt ("Receive each xth line", x); if (p1 != p2) { if (p1 < p2) { for ( i = SelLen-1; i >= 0; i--) { if (i/x != Math.round (i/x) || i/x == 0) { Sel.selected = false; //to remove, change this line with next line //Sel.remove(); redraw(); } } } else { for ( i = 0; i <= SelLen-1; i++) { if (i/x != Math.round (i/x) || i/x == 0) { var j= SelLen-1 - i; Sel .selected = false; //to remove, change this line with next line //Sel .remove(); redraw(); } } } } else { alert("only for x-axis") } Have fun

I already suspected. Thanks for your attempt.

I'll try it tomorrow again. At home I only have CS3.

CS6 is brand new in our company since yesterday. I hope so much that I'm going to find a solution. Because usually I always use a placeholder text in edittext-box in most scripts. Bad news for me ... Probably textselection property is not really the tool of choice for me.

In earlier versions of Illustrator, the text of an edittext-box in a dialog box was automatically selected and highlighted by edittext.active = true; Now in CS6 is only the cursor in text box - no highlighting more. How can I restore the "original state" (javascript in Illustrator CS6, Mac OS and Windows)?

Try this on a single textFrame with numbers and following Letter: #target Illustrator // RoundNumBeforeLetter.jsx // http://forums.adobe.com/thread/1095744?tstart=0 // rundet Zahlen vor einem Buchstaben var myDoc = app.activeDocument; var myTxt = myDoc.textFrames[0]; //the first TextFrame in document var myTxtCharLen = myTxt.characters.length; var myTxtNum, j, LastChar = null; for (i=0; i <=myTxtCharLen-1;i++) { myTxtChar = myTxt.characters.contents; if (SearchText(myTxtChar) == true) { j = i; myTxtNum = myTxt.contents.slice (0,j); myTxtNum = Math.round (myTxtNum); LastChar = myTxt.contents.slice (j); myTxt.contents = myTxtNum+LastChar; } } function SearchText(x) { var myRegExp = /(^-?[A-Z]$)/; return myRegExp.test(x); } Maybe it's not as elegant. On the contrary, it probably is "complicated and a long way around," but should work.

Thanks for the great tuition. CarlosCanto wrote: … phew!! that was hard, grammar proved to be more difficult than programming!! hahaha Now you know how I feel. I learned „my English“ 25 years ago and since then no longer needed. Scripting is indeed easier.

Yes, I agree completely with you. That was exactly, what I would like saying before. (Or: That was exactly what I want to say before.) ??? Do you know, what I mean?

It's difficult to "break" graph with script, because you get grouped grouped grouped items. IMHO the problem will be to select the tick marks with script. To deselect or remove 9 of 10 (or of 100) lines you can give this script a chance: #target Illustrator // Remove9of10.jsx // graph must be breaked; // only if category axis is the x-axis; // all selected tick marks must be ungrouped; //the first tick mark (on position 0 in graph) is also deselected/removed //by pixxxelschubser var myDoc = app.activeDocument; var Sel = myDoc.selection; var SelLen = Sel.length; var p1 = Sel[0].position[0]; var p2 = Sel[SelLen-1].position[0]; if (p1 != p2) { if (p1 < p2) { for ( i = SelLen-1; i >= 0; i--) { if (i/10 != Math.round (i/10) || i/10 == 0) { Sel.selected = false; //to remove, change this line with next line //Sel.remove(); redraw(); } } } else { for ( i = 0; i <= SelLen-1; i++) { if (i/10 != Math.round (i/10) || i/10 == 0) { var j= SelLen-1 - i; Sel .selected = false; //to remove, change this line with next line //Sel .remove(); redraw(); } } } } else { alert("only for x-axis") }

Thanx for your example-file. But what I really need is a file with numbers before and after using script. (Please CS3-conform)

Please show some examples of your numbers and a screenshot with visible layers palette..