Azure：如何找到容器注册表的所有者或资源 ID？

Question

Background

A year or two ago, when I was learning Azure, I created a container registry (ACR).

I would now like to use that exact same name again. I can either use that same registry, or I would like to delete that registry and start again. (It does not have anything useful on it that I do not have in other locations.)

But I cannot seem to figure out which subscription / resource group owns that registry. I am nearly certain I know which one it was, but when I log on to that account, the registry is not there. To be clear, I tried to create it again but it said that the name was not available. It's a very weird name, so I am pretty sure it is still my old one sitting around somewhere.

Question

• How can I find out the owner of a container registry?

In my case, it is a dumb mistake. But I have to imagine that companies will want a particular name, and even be willing to pay a little to have a particular registry name. Is there no way to query Azure to ask for the owner of a container registry? What happens if I lost the "paperwork" for which subscription owns it, like in my case?

I was hoping for something like az acr public show or something like that.

Answer 1

I was hoping for something like az acr public show or something like that

There is a CLI command like it:

az acr show --name acr_name

You can get all the information about your ACR, int includes the owner resource group and its resource Id. The resource Id will show you the subscription ID.

But when you use the CLI, it means you already login and uses the default subscription or set the current subscription. If you don't know which subscription the ACR in, you need to try this command in all the subscriptions your account has.

And for the owner of the ACR, you need to list all the assignment and filter with your requirements:

az role assignment list --role Owner --scope acr_resourceId

This CLI command can list all the owners directly to your ACR, but the list does not include the owners that inherit from the subscription.