Auto log out of Firefox Account when closing
I'm logged into Firefox with a Firefox Account and want to have to log into that account every time I open Firefox after closing it.
When I say "Firefox Account", I mean the one found under "Sync" in Firefox options, or the one managed from accounts.firefox.com.
In Firefox's history settings, I have "Use custom settings for history" checked and only "Clear history when Firefox closes" checked. I have checked "Delete cookies and site data when Firefox is closed."
It seems very strange to me that Firefox would keep me logged into my Firefox Account with this setup.
Additional System Details
- User Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:71.0) Gecko/20100101 Firefox/71.0
Per Wesley in this thread: Unfortunately the ability to delete local data when disconnecting from Firefox Sync was removed in Firefox 71 (see bug 1582023). Too many users were misunderstanding the confirmation screen and accidentally deleting all of their data.
Modified by evhenry
Thanks for the reply. I understand that UX decision but there's still a feature gap here in my opinion.
Anyways, is there some local file(s) I can script to delete myself that would have the effect of logging me out?
Firefox stores the Firefox Account credentials used for Sync in the Password Manager, so if this data isn't stored there then you need to login manually each time you start Firefox.
Once you are logged in to Sync the signedInUser.json file in the profile folder is used to connect to Sync during the current session. I do not know what effect it has if you would remove this file.
The selected answer doesn't actually answer the question. The question is how to auto-logout of firefox sync when a user quits firefox (which also happens to be the question I have).
The selected answer responds with "Unfortunately the ability to delete local data when disconnecting from Firefox Sync..." -- but the question does not ask how to delete local data (and in my case, I don't care about that).
Perhaps I need to create a new post re-asking this question.
Modified by John