Forum Discussion
Get list of attendees on a Teams Meeting programmatically, while meeting is in progress
Hello,
I am trying to programmatically (through C#) get a list of attendees, WHILE the Teams Meeting is in progress. I am the owner of the meeting. I wanted to confirm that I was on the right path, as the sample code that I wrote is not returning anything:
- Is this the right documentation for the above scenario? List participants - Microsoft Graph v1.0 | Microsoft Docs. Specifically, is the following the right call to be used?
GraphServiceClient graphClient = new GraphServiceClient( authProvider );
var participants = await graphClient.Communications.Calls["{call-id}"].Participants .Request() .GetAsync();
- Where can I find the [call-id] of the meeting, which is a parameter that needs to be passed in?
- Will the above code snippet work only when the meeting is in session or can I test after meeting as well?
Sayali-MSFTMicrosoft
abv75 - We are looking into this I will get back to you soon.
Meghana-MSFTabv75 - The same issue is being tracker here - Getting a list of attendees programmatically, during the Teams Meeting (from Command Line Application) - Microsoft Q&A. Please follow this thread.
Hello abv75 I see that you may have posted this question on the answers.microsoft.com site - does the response work? Getting a list of attendees programmatically, during the Teams Meeting - Microsoft Community
I can tell you that when I'm not the organizer of the meeting and I need an immediate list of members, I screen capture the names in the attendee box. Not elegant, but effective.
- Hi,
Thank you for your response. Yes, it is the same question and I posted it here because I was also instructed to post this question on the Teams Developer channel. At this time, I am not sure if my approach is correct and hence wanted some validation from the Teams Developer channel, but I can also try cross posting on the Microsoft Graph channel.
For what I am trying to achieve, I do need to get the information programmatically, not manually.
