Ejecting iPhone from Finder with Applescript or shell

My AppleScript skills are pretty rusty, but if not opposed to using a GUI script you could try:

tell application "Finder"
	activate
	if (count windows) is 0 then make new Finder window
end tell
delay 3
tell application "System Events"
	tell process "Finder"
		try
			--change the row number below for your situation
			click button "Eject" of UI element 1 of row 18 of outline 1 of scroll area 1 of splitter group 1 of window 1
		end try
	end tell
end tell
delay 3
tell application "Finder"
	close window 1
end tell

The script assumes the Finder sidebar items you've chosen to show are static. To get the row number to use in the script, with iPhone attached, count the rows of text, including sidebar headings (e.g., Favorites, iCloud, and Locations), from the top to the row with iPhone. The delays are there so you can see if the script is working. You might try the script with reduced delays or eliminate them altogether. The row the iPhone shows up in Finder's sidebar when connected should (maybe?) be consistent. At least it is on my Mac.

pauclaude wrote:

Hello,

I need to find a way to eject the iPhone (in Big Sur) via a script or a shell command.

The reason is that when I turn off my Mac, very often prevents me from doing it because it is synchronizing the iPhone, and I would like to interrupt synchronization, which can be interrupted only by expelling the iPhone.

In iTunes it was easy to do it, but now the iPhone is managed by the Finder, so I don't know how to do it,

Some solutions?

You can eject the iphone directly from Finder like any other external drive

Use the Finder to sync your iPhone, iPad, or iPod with your ...

Finder>Sidebar>Locations> iOS device

I took a quick look at this. I couldn't see anywhere that the phone was exposed to AppleScript - the Finder didn't acknowledge it as any device that it could reference. As such I can't see any (AppleScript) way to do this.