1

We have a windows server 2012 to which multiple users connect simultaneously and use multiple instances of our application running in different user sessions. The problem is that we have to login each user with Remote Desktop and if RDP is disconnected sessions get disconnected. We need to start and keep the session running without RDP.

Previously in a single user setup on windows 7 we solved it with Auto Logon. Upon login Splashtop or Teamviewer start running and user could connect to system with any of these.

I have setup Auto Login using this method but this logs in only the last configured user with this method. We want to keep multiple logins active and run our apps without using RDP. So my question is how do I auto login multiple users on Windows Server 2012? or alternatively is there a way to keep multiple user sessions active without using Remote Desktop?

Related: Auto-login in Windows Server 2008 (single user only)

1

Enable Multiple RDP Sessions

Log into the server using Remote Desktop.

Open the start screen (press the Windows key) and type gpedit.msc and open it

Go to Computer Configuration > Administrative Templates > Windows Components > Remote Desktop Services > Remote Desktop Session Host > Connections.

Set Restrict Remote Desktop Services user to a single Remote Desktop Services session to Disabled.

Double click Limit number of connections and set the RD Maximum Connections allowed to 999999.

But I fear how all your application supports multiple session. Terminal server or VDI is a complete solution which need to be additional cost.

  • 1
    I have done that already and any number of users can connect to the server using RDP. I need multiple sessions active/login without RDP. – LifeH2O Mar 31 '16 at 11:15
  • This simply does not login multiple users automatically. – LifeH2O Apr 12 '16 at 14:31

Your Answer

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

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