Activate a window using its Window ID

Question

How would I programmatically activate i.e move-to-front-and-focus a window on macOS (not belonging to my app) given its Window ID. My app would run with user grante

Answer 1

I didn't find a way to switch to a specific window yet, but you can switch to the app that contains a specific window using this function:

func switchToApp(withWindow windowNumber: Int32) {
    let options = CGWindowListOption(arrayLiteral: CGWindowListOption.excludeDesktopElements, CGWindowListOption.optionOnScreenOnly)
    let windowListInfo = CGWindowListCopyWindowInfo(options, CGWindowID(0))
    guard let infoList = windowListInfo as NSArray? as? [[String: AnyObject]] else { return }
    if let window = infoList.first(where: { ($0["kCGWindowNumber"] as? Int32) == windowNumber}), let pid = window["kCGWindowOwnerPID"] as? Int32 {
        let app = NSRunningApplication(processIdentifier: pid)
        app?.activate(options: .activateIgnoringOtherApps)
    }
}

It is probably usefull to switch by name as well:

func switchToApp(named windowOwnerName: String) {
    let options = CGWindowListOption(arrayLiteral: CGWindowListOption.excludeDesktopElements, CGWindowListOption.optionOnScreenOnly)
    let windowListInfo = CGWindowListCopyWindowInfo(options, CGWindowID(0))
    guard let infoList = windowListInfo as NSArray? as? [[String: AnyObject]] else { return }

    if let window = infoList.first(where: { ($0["kCGWindowOwnerName"] as? String) == windowOwnerName}), let pid = window["kCGWindowOwnerPID"] as? Int32 {
        let app = NSRunningApplication(processIdentifier: pid)
        app?.activate(options: .activateIgnoringOtherApps)
    }
}

Example: switchToApp(named: "OpenOffice")

On my mac OpenOffice was started with a window with kCGWindowNumber = 599, so this has the same effect: switchToApp(withWindow: 599)

As far as I found out so far, your options seem to be to show the currently active window of the app, or to show all windows (using .activateAllWindows as activation option)