start script from another script

Forum|Forum|1 year ago
April 16, 2025
2 replies
180 views

Hello, everybody!

I have 2 scripts: first.jsx & second.jsx

In UI in first.jsx I have a Button. If I press to this button, I need to start second.jsx.

How can I start second.jsx? Both scripts are in same folder.

first.jsx:

var dlg = new Window("dialog", "Try to start second.jsx");
dlg.testButton = dlg.add("Button", undefined, "start second.jsx");

dlg.testButton.onClick = function(){
    //missing code
}

dlg.show()

second.jsx:

alert("second.jsx successfully started")

Actions and scripting

Stephen Marsh

Community Expert

Similar ways also documented here:

https://gist.github.com/MarshySwamp/eb36460ae2813b6fc47e2d3fec48deb3

Alex RezvovAuthor

Participating Frequently

Now, I try this code. It work.

var dlg = new Window("dialog", "Try to start second.jsx");
dlg.testButton = dlg.add("Button", undefined, "start second.jsx");

dlg.testButton.onClick = function(){
    var curScr = $.fileName;
    var path = curScr.slice(0, curScr.lastIndexOf("/")+1);
    $.evalFile(path + "second.jsx")
}

dlg.show()

Alex RezvovAuthor

Participating Frequently

I think, this code is correct. In this code we don't check existence of "second.jsx":

var dlg = new Window("dialog", "Try to start second.jsx");
dlg.testButton = dlg.add("Button", undefined, "start second.jsx");

dlg.testButton.onClick = function(){
    var curScr = $.fileName;
    var path = File(curScr).parent;
    $.evalFile(path + "/" + "second.jsx")
}

dlg.show()

E

ExUSA

Legend

You can get to the enclosing folder with this snippet:

var x = File($.fileName);
var y = Folder(x.parent);

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