Is it possible to filter users by group (memberOf) when using API endpoint /users to get a list of users?

Question

e.g. https://graph.microsoft.com/v1.0/users?$filter=memberOf/any(c:startswith(c/displayName,+'GROUPNAME')) does not work and we get an error:

{
    "error": {
        "code": "Request_UnsupportedQuery",
        "message": "Unsupported or invalid query filter clause specified for property 'memberOf' of resource 'User'.",
        "innerError": {
            "date": "2023-06-15T11:51:25",
            "request-id": "8fc5b8bd-7e3f-4deb-9054-b7d88110febb",
            "client-request-id": "e54c1f88-9325-a98b-7fc0-6707ef9f2b31"
        }
    }
}

But we need the filter to limit the amount of users which should be returned. Is there any possibility to do this (and if yes, what is the right syntax for the query parameter 'filter')?

Answer 1

Hello Franziska Blume,

Thanks for reaching out. The /users query does not return the membersOf property by default and you have to use the expand parameter for example /users$expand=memberOf query to return group membership of users. Unfortunately, the $expand parameter does not support advanced filtering as mentioned in the documentation. Hence you cannot filter further using the $filter parameter along with the $expand parameter.

---

I would suggest voicing your query and feedback on the Graph Feedback Forum.

You can however run the below query on a user account:

https://graph.microsoft.com/beta/users/[UserId]/memberOf?$filter=startsWith(displayName,'[GroupName]')&$count=true

Please ensure you add the ConistencyLevel to 'eventual' in the header before running the query.

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Thanks.

Answer 2

Hello Franziska Blume,

Good Day. Hope all is well.

If the answer is helpful, please click "Accept Answer" and kindly upvote it. If you have extra questions about this answer, please click "Comment".

Thanks.

Answer 3

Hello Franziska Blume,

Good Day. Hope all is well.

If the answer is helpful, please click "Accept Answer" and kindly upvote it. If you have extra questions about this answer, please click "Comment".

Thanks.