0

I have 2 locations and each has its own WSFC.

  1. Cluster 1: WSFC,Datacenter1, Node1, Node2. Listener1

  2. Cluster 2: WSFC, Datacenter2, Node3, Node4. Listener2

Created an Availability group as EXTERNAL cluster type.

screenshot of AG options in SSMS with "EXTERNAL" selected as the cluster type, which is circled

When simulating Failover, by switching nodes, puts database in RESOLVING mode.

screenshot of SSMS object explorer showing the DB in question is not synchronizing, and one of the AG replicas is resolving

Looking into SQL Error Log on primary replica:

The state of the local availability replica in availability group 'AG_AW2017' has changed from 'NOT_AVAILABLE' to 'RESOLVING_NORMAL'. The state changed because the local instance of SQL Server is starting up. For more information, see the SQL Server error log or cluster log. If this is a Windows Server Failover Clustering (WSFC) availability group, you can also see the WSFC management console.

Looking into SQL Error Log on secondary replica:

A connection timeout has occurred on a previously established connection to availability replica 'Primary' with id [1B93C7DC-75E3-4180-9748-AC1B662781A9]. Either a networking or a firewall issue exists or the availability replica has transitioned to the resolving role

If I try to use WSFC as cluster type, then I will not be able to add Replica. I will get an error saying:

Cannot connect to Listener2. The specified instance of SQL Server is not part of the same Windows Server Failover Cluster (WSFC) as the primary node.

Spent quite bit of time on it, but still couldn't figure out what exactly the problem here.

Improve this question
2
  • 1
    You'll need to provide more details to get help on this, I think. If all of four replicas need to be in the same AG, you shouldn't have two different WSFC clusters and two different listeners. It looks like you might want a distributed availability group (where node1 and node2 are in an AG (AG1), node3 and node4 are in a different AG (AG2), and AG1 is replicating all of its changes to AG2. Is that what you're trying to set up?
    – Josh Darnell
    Jun 29, 2021 at 12:39
  • Thank you Josh. Yes, I think that is what I need. Already gathering info and trying to see how to break existing FCI cluster.
    – Serdia
    Jun 29, 2021 at 13:42

2 Answers 2

Reset to default
1

From the Microsoft Docs for CREATE AVAILABILITY GROUP:

CLUSTER_TYPE

Introduced in SQL Server 2017. Used to identify if the availability group is on a Windows Server Failover Cluster (WSFC). Set to WSFC when availability group is on a failover cluster instance on a Windows Server failover cluster. Set to EXTERNAL when the cluster is managed by a cluster manager that is not a Windows Server failover cluster, like Linux Pacemaker. Set to NONE when availability group not using WSFC for cluster coordination. For example, when an availability group includes Linux servers with no cluster manager.

Since you mention you're using Windows Server Failover Clustering, the implication is that you are using Windows, and should not have the "Cluster Type" set to EXTERNAL. It should be set to WSFC.

Improve this answer
0
1

What you are wanting sounds more like Always On distributed availability groups.

To create a distributed availability group, you must create two availability groups each with its own listener. You then combine these availability groups into a distributed availability group. The following steps provide a basic example in Transact-SQL. This example doesn't cover all of the details of creating availability groups and listeners; instead, it focuses on highlighting the key requirements.

From the Microsoft Docs for Configure an Always On distributed availability group:

For a technical overview of distributed availability groups, see Distributed availability groups.

Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.