How to determine if starting inside a windows service?

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半阙折子戏
半阙折子戏 2021-02-15 16:37

Currently I\'m checking it in the following way:

if (Environment.UserInteractive)
    Application.Run(new ServiceControllerForm(service));
else
    ServiceBase.R         


        
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  • 2021-02-15 17:19

    Why not just use a command line switch?

    // Note that you have to add the params argument, 
    // which isn't usually present in windows services
    private static void Main(params string[] parameters)
    {
        ....
    
        if (parameters.Length > 0)
        {
            if (parameters[0].ToLower() == "/console")
            {
                Application.Run(new ServiceControllerForm(service));  
            {
            else
            {
                ServiceBase.Run(windowsService);
            }
        }
    }
    
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  • 2021-02-15 17:20

    Instead of using the Environment.UserInteractive property, modify the startup method of your service to check for a "-console" command line argument. If the argument is present, then run as an ordinary application. If not, run as a service. It's not as automated as the property check, but it'd be easy to add a shortcut to the desktop that adds the "-console" command line argument for you.

    As an aside, you need to be aware that interaction with the desktop has been disabled in Windows Vista and beyond. If you are running a Windows service that needs to interact with the user, the approved way of doing this now is to separate your front-end application from your Windows service and have them communicate using something like WCF.

    If you need to debug your Windows service (whether it's running as a service or as an application), put a call to System.Diagnostics.Debugging.Break() in your startup method. This will force a prompt that allows you to enter a debugging session. I use this technique to debug my Windows service all the time.

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  • 2021-02-15 17:22

    It's not perfect, but you could probably do something like this:

    public static bool IsService()
    {
        ServiceController sc = new ServiceController("MyApplication");
        return sc.Status == ServiceControllerStatus.StartPending;
    }
    

    The idea is that if you run this while your service is still starting up then it will always be in the pending state. If the service isn't installed at all then the method will always return false. It will only fail in the very unlikely corner case that the service is starting and somebody is trying to start it as an application at the same time.

    I don't love this answer but I think it is probably the best you can do. Realistically it's not a very good idea to allow the same application to run in either service or application mode - in the long run it will be easier if you abstract all of the common functionality into a class library and just create a separate service app. But if for some reason you really really need to have your cake and eat it too, you could probably combine the IsService method above with Environment.UserInteractive to get the correct answer almost all of the time.

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  • 2021-02-15 17:31

    Also, it must be noted, that Environment.UserInteractive always returns true in .NET Core, even if it is running as a Windows Service.

    For the time being, best method seems to be this one from ASP.NET Core.

    Sources: .NET Core 2.2 .NET Core 3.1

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  • 2021-02-15 17:32

    You can check if the process or any of its parent processes are listed as a service:

           var process = System.Diagnostics.Process.GetCurrentProcess();
           var parent = process.Parent();
           var procIsService = process?.IsService;
           var parentIsService = parent?.IsService;
           ...
    
    public static class Extensions
    {
        private static string FindIndexedProcessName(int pid)
        {
            var processName = Process.GetProcessById(pid).ProcessName;
            var processesByName = Process.GetProcessesByName(processName);
            string processIndexdName = null;
    
            for (var index = 0; index < processesByName.Length; index++)
            {
                processIndexdName = index == 0 ? processName : processName + "#" + index;
                var processId = new PerformanceCounter("Process", "ID Process", processIndexdName);
                if ((int)processId.NextValue() == pid)
                {
                    return processIndexdName;
                }
            }
    
            return processIndexdName;
        }
    
        private static Process FindPidFromIndexedProcessName(string indexedProcessName)
        {
            var parentId = new PerformanceCounter("Process", "Creating Process ID", indexedProcessName);
            return Process.GetProcessById((int)parentId.NextValue());
        }
    
        public static Process Parent(this Process process)
        {
            return FindPidFromIndexedProcessName(FindIndexedProcessName(process.Id));
        }
    
        public static bool IsService(this Process process)
        {
            using (ManagementObjectSearcher Searcher = new ManagementObjectSearcher(
            "SELECT * FROM Win32_Service WHERE ProcessId =" + "\"" + process.Id + "\""))
            {
                foreach (ManagementObject service in Searcher.Get())
                    return true;
            }
            return false;
        }
    }
    
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  • 2021-02-15 17:35

    The issue with the accepted answer is that checking the status of an service that isn't installed will throw. The IsService Method I'm using looks like this:

        private bool IsService(string name)
        {
            if (!Environment.UserInteractive) return true;
            System.ServiceProcess.ServiceController sc = new System.ServiceProcess.ServiceController(name);
            try
            {
                return sc.Status == System.ServiceProcess.ServiceControllerStatus.StartPending;
            }
            catch(InvalidOperationException)
            {
                return false;
            }
        }
    

    Which should work more reliably than just checking Environment.UserInteractive

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