User name
I have been using [edited] @comcast.net as user name for Firefox account. I want to delete that email because I'm leaving comcast ASAP. I read instruction that say add a new email name as a secondary account and then delete the primary one above. I entered [edited[ @gmail.com. I got a message that said that account already exists. Since I didn't know that or remember any password for it, I entered [edited] @gmail.com and said "forgot password." I changed it Now Firefox won't let me in using the [edited] @gmail.com account. Would appreciate any help you can offer.
由James于修改
所有回复 (4)
Hi, just to clarify a couple things:
(1) After you added the new Gmail address to your Mozilla Account as a secondary address, did you get the message in your Comcast mailbox to approve the change? And that worked okay?
(2) After you added the new Gmail address to your Mozilla Account as a secondary address, did you switch it to primary? If not, please pause on that until the Gmail side is sorted out.
(3) What happens if you try to sign in to your Mozilla Account using your Comcast address?
Thank you for responding. I already had two accounts, [edited] @comcast.net and [edited] @gmail. I'm going to use the gmail account and want to delete the comcast account altogether. When I go into my profile I'm not sure how to delete the primary email and make the gmail primary.
由James于修改
Hi Sandra, after you sign in to your Mozilla account -- can you get in now? -- there is a two-step dance of adding a secondary address, and then switching the secondary and primary. The following article has more info:
You wrote: I already had two accounts, [edited] @comcast.net and [edited] @gmail
If you already have a @gmail account account then you can't use this email to be used as a secondary email, in that case you can keep using the @gmail account and delete the @comcast account. I'm not sure if the can reuse the gmail account as a secondary account if you would remove the gmail account or that Mozilla/Firefox still remembers that it has been used.