2

I"m having a problem creating a new Availability group in a windows cluster. I have 3 nodes in the cluster. Two nodes are running as a FCI and a single stand-alone node that's a member of the windows cluster.

Msg 19405, Level 16, State 17, Line 2 Failed to create, join or add replica to availability group 'SQL Cluster 2', because node 'DRNode' is a possible owner for both replica 'DRNode' and 'SQL Instance'. If one replica is failover cluster instance, remove the overlapped node from its possible owners and try again.

I've fixed this issue by removing the DRNode from being a possible owner of the SQL FCI, but the issue persists.

Do I need to restart the role to make it effective?

I've confirmed in PS that it's correct.

 PS C:\Windows\system32> get-clusterownernode -resource "SQL Network
 Name (FCI Name)"

ClusterObject                                               OwnerNodes
-------------                                               ---------- 
SQL Network Name (FCI Name)                                 {Node1, Node2}
  • check this: mssqltips.com/sqlservertip/4967/… – CR241 Mar 21 at 23:27
  • 1
    The PS output shows the network name is not able to be owned by Node3, but what about the SQL Server instance resource itself? – HandyD Mar 22 at 3:56
  • It's the same as the network name. The weird thing is that I have an identical cluster that's working fine, so I have something to compare it to. And that SQL instance is able to be owned by all 3 nodes. But on the problem cluster, the 3rd node is not a possible owner, and even if I select it, click ok, then reopen it, it's unselected again. – Dmitry Mar 22 at 4:25
  • CR241 - Yes I've seen that, I have all the symptoms, but I've implemented the solution they gave but the issue persists. I"ve even evicted the 3rd node then re-added it so try to clear out any weird lingering issue but the problem persists. – Dmitry Mar 22 at 4:27
  • Can you provide a list of all ClusterObjects and their respective OwnerNodes? – John Eisbrener Mar 22 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.