Swift - how to make window unactivable?

Question

I want to make application unactivable.

I mean its must be inactive and non-foreground even by click on it. Active window must be the same app as it was before click on my app.

How can I do this?

---

upd:

1. Let's imagine some window/panel. Like a Dock. Let's call it "Docky"

2. Docky is inactive, Active window is Safari:

   NSRunningApplicationSafari.isActive == true
   NSRunningApplicationDocky.isActive == false
   

3. I'm clicking on ANY empty point of Docky.

Expected result:

4. Safari is still active window. Like before click:

   NSRunningApplicationSafari.isActive == true
   NSRunningApplicationDocky.isActive == false
   

Actual result:

4. Docky is active window, Safari is inactive:

   NSRunningApplicationSafari.isActive == false
   NSRunningApplicationDocky.isActive == true
   

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Another sample is Keyboard Viewer. You are clicking on the virtual keyboard, but active window is another app. Ant exactly active window getting keyPress events.

Answer 1

Actually all you need is .nonactivatingPanel style panel. Everything else is details, like level of this window, custom views with overridden acceptsFirstMouse:, needsPanelToBecomeKey, etc. Btw, button accepts first click by default, non activating app in this case.

So your AppDelegate, for example, might look like the following:

class AppDelegate: NSObject, NSApplicationDelegate {

    var docky: NSPanel!

    func applicationDidFinishLaunching(_ aNotification: Notification) {

        // Create the window and set the content view. 
        docky = NSPanel(
            contentRect: NSRect(x: 0, y: 0, width: 120, height: 600),
            styleMask: [.nonactivatingPanel],
            backing: .buffered, defer: false)
        docky.level = .mainMenu
        ....
        docky.orderFrontRegardless()
    }