How can I search for users in repo?

[Fricco]

Hello, as we are having several contractors leaving the company, I need to review their access to our repo and I can't find an easy way to search if a particular user has access or not. What I have been doing is manually scrolling through the list of more than 100 developers to see if they have access to it or not. 

I would like to know if there a simple way of doing this, basically if I can search of a user within the repo and if found to be able to remove them.

When going to User Groups section I see there is something that looks like a search box, but is only to be able to add new users, not so search within the list of existing ones.

Thank you.

[Stephen Sifers]

Hello Francisco,

That task you’re attempting to accomplish can be done from the “Users on plan” screen which will display all users which are members of any and all repositories. To get the “Users on plan” screen, follow these steps:

1. Select your avatar in the bottom left and click Bitbucket Settings
2. Within Setting and under Plans and Billing select “Users on plan”
3. Within this list, find the users which you want to remove from Bitbucket all together and select the X to the far right.
4. You will be prompted to confirm you’re about to delete a user and you will also see all repositories the user had access to.

Source documentation: Delete an account | See: Deactivate or delete managed accounts

If you’re still having issues finding and removing users, please do let us know.

Regards,
Stephen Sifers

[Fricco]

Hi Stephen, this still doesn't allow me to search for specific users. We have 210 users and I have to navigate through pages and pages to find out a particular user. Still your option shows a larger list meaning less pages to navigate.

[neil]

Hi @Stephen Sifers is this going to be added as a feature? It's a bit ridiculous that I have to navigate through 27 pages of users just to find someone's account to delete them from the plan. The solution you've provided doesn't answer the question. A search function is absolutely necessary here.