SOLVED

Wildcards in Graph to see all planners?

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Occasional Contributor

Hi Guys

 

I am quite new to Graph and I was wondering if there is functionality I am missing. I am sure I can write some very long PS script to do this but worth a post before I start...

 

So I want to see all the Planners in my Organisation, I know I can use graph to retrieve all the Plans associated with a group: https://graph.microsoft.com/v1.0/groups/<group id>/Planner/Plans

However there seem to be two problems with this:

 

  • I have to give myself access to the group to be able to query it, despite graph having perms to read group etc.
  • I have to do it one group at a time.

 

So before I start writing laborious scripts and adding myself to all 365 groups, is there anyway to deal with these two issues in a more fluid way. I.e. get access (i'm a global admin on the tenancy) and then list all the groups in one query - maybe a wildcard?

 

I am asking here as I can't find anything out there which seems to answer these issues.

 

Cheers
Dan

2 Replies
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Solution

@danensor 

 

To my knowledge, there is no way to get plans and its data without being a member of the plan's associated group even if you have global admin privilege.

 

If you have added yourself as member in all plans (groups),  then you can use below end-point to lists all the plans where you are member, but this api also has some problem as it will list plan only when you personally visit the plan in Planner portal at-least onetime. To my knowledge, both the api end-point and the issue is not documented anywhere.

 

# Not recommended to use this api without proper testing.
https://graph.microsoft.com/v1.0/me/planner/plans

 

With Global Admin privilege what you can do now is : You can export specific user's all plans data without being a member of the user's plans. You can refer below post for more details.

 

https://support.office.com/en-us/article/export-user-data-from-microsoft-planner-91258c96-b353-4da1-...

 

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@kombaiah Murugan That's a great idea, thanks!