Sip and Ring simultaneous

Jami would be a perfect solution for my project if it could register on a sip server aswell as be registered to the ring network… This would enable a sip initaited caller to reach an app user via sip albeit with limited functionality but the call would go through and vice versa. Would want sip as priority to try an initate a call and if routing fails after x time, app looks up ring network directory to place the call.

Can this be done (create the coding)?

Merci a plus
Christophe

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I assumed this was a given. Thanks! Now I know I need a separate SIP client.
That would be great, to have bot simultaneously.

Hi @mickeymouse690 You mean you want to link SIP account contacts with Jami account contacts so you can default to one or the other?

You can already can create an unlimited number of SIP and/or Jami accounts. Jami can make SIP calls through a server as well as normal P2P SIP calls routed using the Jami network.

No, not linking contacts. I just assumed I could make calls to either phone numbers or jami users. For example, I want to be able to call for a doctor appointment and then simply call my jami friends without needing to reconfigure anything. I assumed that was possible. Did I misunderstand?
Actually, I get it now. I just need to switch accounts. But can I still receive calls from either account?
BTW, I;m trying to make a jami/sip phone on a raspberry, to show kids that anyone can do it. I found this jami app, it seems ideal.
Edit: kids means “college kids”, I;m that old!

it’s already possible and supported by client.
If you have a current SIP call and a current Jami call you can merge them via the add participant button. But it’s two separate accounts

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Thanks! How about for incoming calls? Can you receive for multiple accounts, or is it only the one that is selected?

you can receive from multiple accounts (the ui is connected to different accounts at the same time) then merge manually the calls if necessary.

But if someone call an account only the account receiving the call can answer

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