May I join an already existing chat room (e.g. the Logic one) and invite another user to talk ?
If so, how ?
Thanks :-)
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$\begingroup$ See here: How do I invite a user to chat? However in case of room created by another user, it might be polite to know what the room owner thinks about that before inviting other users. $\endgroup$ – Martin Sleziak Jan 12 '18 at 11:21
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1$\begingroup$ You can find link to chat faq and also some other useful links in this tag-info. $\endgroup$ – Martin Sleziak Jan 12 '18 at 11:23
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$\begingroup$ @MartinSleziak - thanks ! The issue was that I was able to create a new room but in both cases I cannot find how to invite someone. At the end (soway) we succeed. :-) $\endgroup$ – Mauro ALLEGRANZA Jan 12 '18 at 12:14
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$\begingroup$ Actually, you are currently an owner of the room! So you're already jointed. See: chat.stackexchange.com/rooms/info/44058/logic?tab=access. And as an owner you can certainly invite other users. $\endgroup$ – Namaste Jan 12 '18 at 15:23
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$\begingroup$ @amWhy - yes thanks ! We succeeded (with the help of some user...). But the issue is that I do not know how to do it. $\endgroup$ – Mauro ALLEGRANZA Jan 12 '18 at 15:26
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What the Chat FAQ says about inviting is interesting but not explicit:
If you need to invite someone to your room, do it from the user's chat profile page or from the user card that pops up when you click on their avatar in a chat room.
I assumed this meant searching for a user from the Chat Users page. Although I drilled down to find myself, this did not immediately suggest to me a mechanism for issuing an invitation.
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$\begingroup$ If you have some problems finding mechanism for issuing an invitation, perhaps the details explained here might help: How do I invite a user to chat? $\endgroup$ – Martin Sleziak Jan 13 '18 at 9:30
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$\begingroup$ Thanks, @Martin : your Answer there points out a subtlety I would not have thought about - if I want to invite a user to an existing Chat, I have to be in that room currently for the invitation option to appear. $\endgroup$ – hardmath Jan 13 '18 at 14:23
