0

I just setup SQL Server 2012 (on Windows Server 2012) Always On feature and have a problem with connecting to SQL Server Listener from non-primary node (it works from primary node though).

Here is the scenario:

2 nodes in different subnets: 
   SQL1 10.0.2.11 (primary)
   SQL2 10.0.6.11 (secondary)
Listener:
  Name: SQLPROD1
  Port: 1433
  Subnets       IP address
  10.0.2.0/24   10.0.2.102
  10.0.6.0/24   10.0.6.102

I can connect to SQLPROD1 in SSMS on primary node (either SQL1 or SQL2) but can't connect (not even ping) SQLPROD1 from secondary or from any other client in the network.

I checked firewall and it doesn't seem to be the cause of the problem. I can ping SQL1 or SQL2 from any client but can't ping SQLPROD1.

Beside this, everything else works in Always On. I can successfully fail-over and see databases on other node.

3
  • 1
    Are you sure that the hostname is resolving to the right IP address, or resolving at all? That would be my first verification. Mar 7 '14 at 18:45
  • 1
    Have you verified the listener is configured for both subnets? Jan 16 '15 at 19:43
  • You using two different network segments, is there any route between them?
    – McNets
    May 9 '19 at 13:00
0

We had the same issues as this. Recreated the DNS entries for the two listeners and making sure "Allow any authenticated user to update DNS records with the same owner name" is checked. Believe the failover event caused issues with the listener not updating it's IP address.

0

Just had a similar issue. All I did was manually failover to secondary node and listener name became pingable again. Since the name is only configured in SQL (and not DNS) it appears SQL lost track of the name for some reason but the failover corrected the problem.

1
  • 1
    SQL server registers the listener name in DNS and AD. The cluster machine account needs to have
    – swasheck
    Jan 2 '19 at 23:54

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .