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Authorizathion new members in a channel

rita87
Listener

We are facing from some weeks a change in adding new members in channel. Now, when I add a new member in channel, although I am able to add everybody, a request is sent to the new member and he/she could not read and be part of the channel unitl when he/she doesn't accept the invitation. Before this change, when I added somebody, he/she was immediately in the chat without any acceptance. Is it possible to avoid this kind of authorization, by changing any settings in the channel? I've tried to do it, but I cannot find any setting to skip this "invitation to be parte of the chat". 

3 REPLIES 3

anallabres
Listener

I am facing the same problem. I would be very grateful if someone can tell us how to avoid that. Thanks!

Hamilton-Zoomie
Participant | Zoom Employee
Participant | Zoom Employee

Sorry folks, as part of the last release and at the guidance of risk and compliance, when inviting external members of Team Chat to a Group Chat or Channel, they get a notification and the option to decide if they want to join it.  In your chat/channel, they show as pending until they accept.  There are lots of good use cases for why that should be in place.
Now having said that, the Team Chat product team has heard loud and clearly from customers that this introduced a decent amount of friction in the process.  They are quickly evaluating a solution that makes customers happy but gives users more control over being added to chats/channels without their consent.
I will update this thread as soon as I have a more clear picture of how the team plans to smooth out this process.

In the mean time, send a DM to those you invite and ask them to accept.

anallabres
Listener

Hi Hamilton-Zoomie,

Thanks for your response. We do a team meeting every week and I sometimes need to remove people from the call for a week and then put them in back again. Is there a way of doing that without bothering people every time to accept an invitation?

Maybe I am just doing it wrong or over complicating and there is an easy way of doing that?

Thanks!

Ana