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H
Heimdaal
Inspiring
September 14, 2016
Answered

Trying to get path of running JSX ($.filePath not working)

  • September 14, 2016
  • 2 replies
  • 6780 views

I'm trying to get the location of the directory of the current script (main.jsx of my panel).
I've tried using $.filePath without success...

alert($.filePath);
Returns undefined or an int

var thisFile = new File($.fileName);
alert(thisFile);

Returns /c/Program%20Files/Adobe/Adobe%20Photoshop%20CC%202015

var thisFile  = new File($.fileName).parent;

alert(thisFile .fsName);

Returns C:\Program Files\Adobe\Adobe Photoshop CC 2015

What am I doing wrong?
I need to get the path of the running script file so I can do .parent to get it's folder name, and from there get a sibling folder name where I have some stuff to retrieve. I'm running this in my main.jsx -file.

This topic has been closed for replies.
Correct answer Heimdaal

Found the solution!

By running this code in the javascript file of your extension you will get the path:

var csInterface = new CSInterface();
var extensionRoot = csInterface.getSystemPath(SystemPath.EXTENSION) + "/jsx/";
alert(extensionRoot);

Credits goes to Davide Barranca

2 replies

H
HeimdaalAuthorCorrect answer
Inspiring
September 19, 2016

Found the solution!

By running this code in the javascript file of your extension you will get the path:

var csInterface = new CSInterface();
var extensionRoot = csInterface.getSystemPath(SystemPath.EXTENSION) + "/jsx/";
alert(extensionRoot);

Credits goes to Davide Barranca

Pedro Cortez Marques
Pedro Cortez Marques
Legend
September 14, 2016

Hi

Using this

File($.fileName).fsName

you'll get  C:\Program Files\Adobe\Adobe Photoshop CC 2015\[scriptName.jsx]

Using this

decodeURI(File($.fileName))

you'll get  /c/Program Files/Adobe/Adobe Photoshop CC 2015/[scriptName.jsx]

Or, to get the folder:

Using this

File($.fileName).parent.fsName

you'll get  C:\Program Files\Adobe\Adobe Photoshop CC 2015

Using this

decodeURI(File($.fileName).parent)

you'll get  /c/Program Files/Adobe/Adobe Photoshop CC 2015

You only need to use decodeURI() if not getting fsName, because fsName result is always decoded.

H
HeimdaalAuthor
Inspiring
September 14, 2016

It's not working.
Using your first example, I get:

C:\Program Files\Adobe\Adobe Photoshop CC 2015\526

then the second time I use it I get

C:\Program Files\Adobe\Adobe Photoshop CC 2015\527

The number increments for each time I run the function where I have alert(File($.fileName).fsName);

Pedro Cortez Marques
Pedro Cortez Marques
Legend
September 14, 2016

Hi

Oh I have noticed just now that you are not putting your scripts in the scripts folder.

They should be here:

C:\Program Files\Adobe\Adobe Photoshop CC 2015\Presets\Scripts\

I have put a script jsx file in the photoshop scripts folder containing this:

alert(File($.fileName).fsName);

and the alert came up with this:

C:\Program Files\Adobe\Adobe Photoshop CC 2015\Presets\Scripts\myScriptFile.jsx

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