How do I configure the name of a Windows service upon installation (or easily at compile time)?

31,096

Solution 1

I tried accessing a configuration using

ConfigurationManager.OpenExeConfiguration(string exePath)

in the installer, but couldn't get it to work.

Instead I decided to use System.Environment.GetCommandLineArgs() in the installer like this:

string[] commandlineArgs = Environment.GetCommandLineArgs();

string servicename;
string servicedisplayname;
ParseServiceNameSwitches(
    commandlineArgs, 
    out servicename, 
    out servicedisplayname);

serviceInstaller.ServiceName = servicename;
serviceInstaller.DisplayName = servicedisplayname;

Now I can install my services using

InstallUtil.exe /i InstallableService.dll /servicename="myserviceinstance_2" /servicedisplayname="My Service Instance 2"

I wrote up a more elaborate explanation here.

Solution 2

You can't pass this in as a command line arg, since InstallUtil doesn't provide the right hooks for that.

However, you can make your service installer read the ServiceName from a config file. If you look at some code for a typical ServiceInstaller, you'll see it's just a matter of having the appropriate DisplayName and ServiceName properties setup at runtime. These could easily be read from a configuration file instead of being hard-coded.

Solution 3

Instead of using Environment.GetCommandLineArgs(); the class Installer has a property called Context from which you can access command line arguments passed to InstallUtil structured in a nice StringDictionary.

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Rune
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Rune

I am a software architect and developer with 15+ years of experience. I have spent most of my career on the Microsoft .NET platform building APIs and integrations. On a day-to-day basis I write most of my code in C#, although I prefer and have done quite a bit of professional work in F# as well. I often do teaching and mentoring, and occasionally give talks on Cloud Architecture, DevOps, or F# (example from F# eXchange 2019 in London: How F# helped us build a more robust application).

Updated on July 09, 2022

Comments

  • Rune
    Rune almost 2 years

    I've created a Windows service in C#, installed it on a server and it is running fine.

    Now I want to install the same service again, but running from a different working directory, having a different config file etc. Thus, I would like to have two (or more) instances of the same service running simultaneously. Initially, this isn't possible since the installer will complain that there's already a service with the given name installed.

    I can overcome this by changing my code, setting the ServiceBase.ServiceName property to a new value, then recompiling and running InstallUtil.exe again. However, I would much prefer if I could set the service name at install-time, i.e. ideally I would do something like

    InstallUtil.exe /i /servicename="MyService Instance 2" MyService.exe

    If this isn't achievable (I very much doubt it), I would like to be able to inject the service name when I build the service. I thought it might be possible to use some sort of build event, use a clever msbuild or nant trick or something along those lines, but I haven't got a clue.

    Any suggestions would be greatly appreciated.

    Thank you for your time.

    • Nate
      Nate almost 15 years
      Is there a reason you haven't given your service the abbility to execute the business logic in n-threads for n-config files? Thus saving the multiple instances issue from the start?
    • Rune
      Rune almost 15 years
      Hmmm, that might be a good point. But the service was created a year ago, way before I realized that I would like to be able to run multiple instances, so the application isn't really architected appropriately. However, I've found a solution and will post it in a sec.
  • Darragh
    Darragh about 13 years
    Really? I've tried reading these values from the ServiceExeName.config file during installation of the windows service and it didn't work :(
  • techExplorer
    techExplorer about 8 years
    Look at question and answer to get perfect solution for this: stackoverflow.com/questions/8516701/…
  • cja
    cja over 4 years
    Please could you post your "more elaborate explanation" somewhere? Your link doesn't work and I'd like to read it