I need to remove a room from specific meeting using EWS (SOAP).
I have an impersonation user and using it, im manage to get a list of all the meetings in a specific room, but I wan't manage so fur to update the meeting by removing the room from it (make the room available for booking).
I tried to remove all attendees from the meeting, it removes the attendees but keeps the room, I tried to remove location - it clears the location field but still keeps the room as meeting attendee.
Help would be appreciated, near to give up here :(
OK, just solved it:)
The answer was to impersonate a room account and then simple decline the invitation.
hope it will help to someone...
Related
I am very new to coding. but i managed to build a group chat app using a low code platform called flutterflow. i managed to spend significant amount of time on it and was able to build a public group chat app except few functionalities. I am hoping to find help from here. for the following questions.
I have chat mods appointed on a group level. like if you create a group, you are a founder and you can assign mods to that perticular group chat. now i want these mods to be able to ban a user in that particular group chat.
I have tried created a subcollection in groups called "banned user" and created two feilds. one is "banned users" document reference to users. and another is "banned_till" to record a time stamp until the user gets banned.
Problem with this is when i ban a user twice, it creates two documents in the user reference with the same user. and two documents has different "banned_till" times. which one it is supposed to pick?
i tried to do this and put a conditional visibility to the chat that "if current time is less than or equal to banned_till time" it wont let user type in the textfield to chat. but this is giving me gray screen.
I am very new to this. any help would be appreciated.
there is specific way to do so. you have to set custom logic. like save all banned users in a firebase database object with thier max time and procced next.
I'm currently organizing the Google groups that our account has and would like to see where each group received the email.
As far as I'm looking at the public API, it seems that there is no way to know when a group received an email or how many it already received.
For example, if a member has been a member of a group since the beginning, it would be possible to create a list from the member's email archive, but I don't want to do that for each group.
Does anyone have a better idea?
Maybe one of the solutions is here, and I'll check it.
Download all messages from a Google group
Could someone give me a link (github/gitlab) of a swift projet which show the possibility of managing Users on Parse.
For example , I would like for a user (who have 15 friends):
to create a group with 5 friends for example (the user could add a name for this group)
to delete if necessary the group
to send a message(textfield) to a group
to send a message (textfield) to a user
For the UI, I would like if it exists, to manager user like the Springboard (having the picture of user in round), and when you longpress on it, you have a cross to delete /and 'block sign' to block user.
And the user could drag/drop other user on his friend list to create a group (like on the Springboard when you create a group of app)
Thanks for helping me !
Have a nice day!
I've not come across such framework for parse/swift. I think you need to build it yourself.
This might help you in the direction:
Friend/de-friend people: https://stackoverflow.com/a/32557977/3314336 &
Parse Swift: User relations with a "friends request"
Send messages framework: https://github.com/slackhq/SlackTextViewController
What you're asking for is pretty general and extensive. You could use Roles and Relations to manage friend groups. You could have Group objects, give them an owner, and a relation for all the members of the group.
For chat, I'd recommend not trying to just build that directly into Parse. Twilio recently released their Programmable Chat feature, which is awesome. Sounds like it'd be beneficial for you to use that.
I am working on the chat application.
I want to implement the group user chat and I have successfully done that.
Now I want toad the functionality by which user can leave the group. I found leave the group but it seems like it's just making user unavailable but not removing user from the member list.
Is there any way by which user remove himself from the member list?
I have read the xep 0045 for it, in that there is a topic for existing the room which seems make user unavailable but not removing from member list.
http://xmpp.org/extensions/xep-0045.html#exit
If anyone has any idea how to do this please share it.
Leaving a room in XMPP (0045) is accomplished by sending an unavailable presence to that room (see XMPPRoom.leaveRoom() in the XMPPFramework) - that should remove them from the occupants of that room (we do this with our app regularly).
"Members list" - is a MUC feature for moderator use cases, it is not related to user. Just leave room as described in XEP, you no need to modify members list. In fact, you no need to add user in this list too, simple chat application should only tracks <presence> stanzas from and to room JID.
There is a way to get the rooms in which you have been invited?
thanks in advance.
No, such information is not stored on the server. You receive an invitation once, as a message - and must handle it then or store it yourself. For the format of invitations, see XEP-0045: Inviting a user to a room.
You can also save rooms you want to remember as "bookmarks", see XEP-0048.