I use Apache Commons Daemon to run as my Java program window service. I want the service to start automatically after installation, i.e.: the startup type will be 'automatically' and the service status will be 'Starved'.
I startup type 'Automatically' ' - startup = auto
), but it does not start automatically, I manually click on the' Start 'button for the first time And only after that it works.
How can I set the status to 'start' without getting the user started manually?
Here is my complete deployment line:
myService.exe // IS // MyService --Install = C: \ MyService \ bin \ myService.exe - Description = "My Java service" --JVM = Auto - ClassPath = C: \ MyService \ classes --StartMode = jvm --StartClass = Test.Service.MyService --StartMethod = windowsService --StartParams = start --StopMode = Jvm --StopClass = Test.Service.MyService --StopMethod = windowsService --StopParams = stop - Logpath = C: \ MyService \ logs --StdOutput = auto --StdError = auto --Startup = auto
Thanks
Resume the solution .. I forgot it , And this is why the service is not starting E.
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