How to get the paths of mailboxes in Apple Mail?

Hi,

I try to collect all IMAP Folders in Apple Mail and I want the complete folder paths. With this code I get all the references to the mailboxes:

tell application "Mail"
	set folders to mailboxes
	repeat with theAccount in accounts
		set folders to folders & mailboxes of theAccount
	end repeat
	return folders
end tell

The entries don’t have a property with the path, but Script Debugger shows me this result. How to get the highlighted part of the objects with AppleScript?

Each mailbox can get parent mailbox.
So, you can go up to root.

Thanks, I already tried it, but building a list of 286 folders is quite slow with 4 seconds. Maybe this can be optimized? I thought it would be faster to access the information which seems to be already there.

tell application "Mail"
	set folders to mailboxes
	repeat with theAccount in accounts
		set folders to folders & (mailboxes of theAccount)
	end repeat
	set allPaths to {}
	repeat with theFolder in folders
		set thePath to ""
		set subFolder to theFolder
		repeat
			try
				set thePath to name of subFolder & "/" & thePath
				set subFolder to container of subFolder
			on error
				exit repeat
			end try
		end repeat
		try
			get name of account of theFolder
		on error
			set thePath to "Local/" & thePath
		end try
		set allPaths to allPaths & {characters 1 thru -2 of thePath as string}
	end repeat
end tell

Execution speed depends on how many messages and mailboxes and the depth of each mailboxes.

My Mail.app env store hundreds of mailing lists.
My M1 Mac mini + macOS 13.6 env’s result with 286 mailboxes:

Your logic (+ little fix): 32.01 sec
My logic : 29.01 sec
Your logic (+ full speed update): 19.32

These scripts calculate full Mail box path from selected mailboxes.

This is not a strict benchmark. But after some times execution, this trend is reasonable for me.

Generally speaking, your script is not so slow.

Thanks for the improvements, Piyomaru. That’s great.

Why does storing the arrays in a script speed up the execution?

Reading huge list take a long time.
This figure is from my book “AppleScript speed up technics”

Direct list access take a long time. And indirect list access is faster than direct.
I don’t know about the mechanism. AppleScript’s array (list) can store various object.
So, its address calculation took a long time, I think.

We experienced this phenomenon in Classic MacOS era (a reference to).
Storing list in script object is new technic in Mac OS X era.

Normal list access script took 15 secs with my M1 Mac mini

use AppleScript version "2.4"
use scripting additions
use framework "Foundation"

property aList : {}

set a1Dat to current application's NSDate's timeIntervalSinceReferenceDate()

set aList to {}
repeat with i from 1 to 30000
	set the end of aList to i
end repeat

repeat with i in aList
	set aStr to i as string
end repeat

set b1Dat to current application's NSDate's timeIntervalSinceReferenceDate()
set c1Dat to (b1Dat - a1Dat)
--> 15.198549985886

Indirect list access (a reference to) took 0.44 secs

use AppleScript version "2.4"
use scripting additions
use framework "Foundation"

property aList : {}

set a1Dat to current application's NSDate's timeIntervalSinceReferenceDate()

set aList_ref to a reference to aList

set aList to {}
repeat with i from 1 to 30000
	set the end of aList_ref to i
end repeat

repeat with i in aList_ref
	set aStr to i as string
end repeat

set b1Dat to current application's NSDate's timeIntervalSinceReferenceDate()
set c1Dat to (b1Dat - a1Dat)
--> 0.439829945564

property in script object is the fastest access. It took 0.39 secs

use AppleScript version "2.4"
use scripting additions
use framework "Foundation"

script hgs
	property aList : {}
end script

set a1Dat to current application's NSDate's timeIntervalSinceReferenceDate()

set (aList of hgs) to {}
repeat with i from 1 to 30000
	set the end of (aList of hgs) to i
end repeat

repeat with i in (aList of hgs)
	set aStr to i as string
end repeat

set b1Dat to current application's NSDate's timeIntervalSinceReferenceDate()
set c1Dat to (b1Dat - a1Dat)
--> 0.38650393486
2 Likes

Thanks a lot for sharing this. Very interesting.

That’s great info, thanks. The speed difference is indeed huge in your example.

Could reproduce the same results on my side, too.

I never knew about it. It also made me wonder why I never experienced any kind of noticeable delay when iterating through lists using the direct access.

I guess I just never had to deal with lists that contain thousands (let alone tens of thousands) of objects.

I ran some additional tests and it looks like that humanly noticeable difference in speed starts from about 4000 objects in your example.