Launch a winform application from a windows service

Question

Hi,

Please let me know how do I run the app under current logged in user from the service.

To give you background, I have a VB.NET Windows service whose sole functionality is to run a Winform App at a specified time. Apart from that it also sets a system wakeup timer so that the system can be woken up at the specified time, if it goes into standby/sleep, to run the app. This service has to cater to XP/Vista/Win7 desktops on our network. This service won't run on servers and laptops.

The Winform App shows a UI for the user to provide some inputs. If the user does not provide the input within 15 minutes, then it defaults the value and then goes into system tray icon. The user can click on the icon and change the values later (within in a specified time frame and that too only twice).

There is absolutely no interaction between the service and the winform app apart from the service starting the app. It also monitors if the app has been killed by the user/crashed. If it has been killed/crashed, then a new instance is run after 30 mins from previous run.

If there is no user logged on, then also I want the app to be run at the specified time. As I said before, the app has a default timer. So if some user has just logged off from the system, then defaults would be set by the winform app.

Now coming to why I am stuck with this design - I cannot use TaskScheduler because it has been disabled on all machines and security team is not willing to change it. TaskScheduler had the option to wakeup the machine from sleep and other things. So basically I ended up creating a service which is acting like task scheduler.

Currently when I run the app.exe via process.start() within the service, its running under SYSTEM account as the service is also running under LOCAL SYSTEM. So basically I am not getting any UI. Is there anyway to run it under the current logged in user? I am not worried about multiple user login as we wont be running it on servers and switch user is not enabled on our desktops. Even if somebody has done a remote login via mstsc, then also I need the run the app and show the UI to the user.

Please let me know how do I run the app under current logged in user from the service.

Thanks askids

Answer 1

There were some additional comments that I posted. But I somehow cannot see it :(

Coming back to the original question. I was able to figure it out after several trial and errors. I will put it in detail.

With Vista and above, services run in isolation from other user sessions. They run in session 0. User sessions run in 1 and above. So basically you need to emulate the process as current logged in user.

1. Use WTSEnumerateSessions and get of sessions. Check if the sesion state is active. This will be current logged on user session. If there are no active sessions, it means there is no logged on user. In my case, there will be only 1 logged on user. So I need not figure out the active session (like others may need to do).
2. Use WTSQueryUserToken to get the user token in the active session. Create a primary user token using DuplicateTokenEx
3. Create an environment using CreateEnvironmentBlock
4. Use the information above in the CreateProcessAsUser

The reason why it was working in XP and not in Vista was because it looks like the startup default information is different. After I set wShowWindow flag of the startupinfo structure, the GUI would start appearing.

Dim StartupInfo As New STARTUPINFO()
StartupInfo.cb = Marshal.SizeOf(StartupInfo)
StartupInfo.dwFlags = STARTF_USESHOWWINDOW 
StartupInfo.wShowWindow = WINDOW_STATUS.SW_SHOWNORMAL

One more additional info. I was trying to set the default desktop using
StartupInfo.lpDesktop = "WinSta0\\Default"
because of which the application would crash upon launch. So I commented it out.

I still have one final issue. The launched app is not in focus. The GUI appears, but in background. But I am thinking, it will once again have to do with some parameters like above. Once I figure it out, I will add in the details.