how to know Office 365 user ever logged in using mobile or desktop or Browser for different workloads like Teams , Exchange, Sharepoint

Venkata Lakshmi Priyanka Gontla 0 Reputation points
2023-08-29T14:58:09.41+00:00

Hello,

I'm currently working on creating a license recommendation tool in powershell to recommend a license to user. I need to check if user is only using mobile for accessing office 365 apps like teams , outlook , sharepoint etc. if he logged in desktop or browser I should get that data as well.

I came across one api to find the data GET https://graph.microsoft.com/v1.0/reports/getM365AppUserDetail(period='D7')?$format=application/json. I am not sure this is giving proper data

Any guidance or examples would be greatly appreciated.

Thank you in advance for your assistance!

Outlook | Windows | Classic Outlook for Windows | For business
Microsoft 365 and Office | Install, redeem, activate | For business | Windows
Microsoft 365 and Office | SharePoint | For business | Windows
Microsoft Teams | Microsoft Teams for business | Other
0 comments No comments
{count} votes

1 answer

Sort by: Most helpful
  1. Vasil Michev 119.7K Reputation points MVP Volunteer Moderator
    2023-08-29T15:37:17.0033333+00:00

    While the "app usage" report can give you some clue, I'd recommend using the Microsoft 365 Active users report instead (getOffice365ActiveUserDetail), which gives you the "last activity date" per workload. Keep in mind that all reports show data with some delay, so it's not exactly current, but for your purpose, it should do.

    Alternatively, you can crawl the Unified audit log entries to determine "last activity date" per user per workload, but that's a lot of overhead compared to leveraging the built-in reports. Still, it's the only way to get proper data for workloads for which we do not have matching Graph API endpoints just yet.

    0 comments No comments

Your answer

Answers can be marked as Accepted Answers by the question author, which helps users to know the answer solved the author's problem.