Relative Path issue with .Net Windows Service..?

I have a windows service which is trying to access an xml file from the Application directory.

Windows Service Installed directory : C:\Services\MyService\MyService.exe

Path of the xml file : C:\Services\MyService\MyService.xml

I am trying to access the file using the following code.

using (FileStream stream = new FileStream("MyService.xml", FileMode.Open, FileAccess.Read))
  {
         //Read file           
  }

I get the following error.

"Can n开发者_如何学Cot find file : C:\WINDOWS\system32\MyService.xml"

My service is running with local system account and I don't want to use absolute path.

---

There is an elegant solution for this from the following link.

http://haacked.com/archive/2004/06/29/current-directory-for-windows-service-is-not-what-you-expect.aspx/

As my service is running both as console/service I just called

Directory.SetCurrentDirectory(AppDomain.CurrentDomain.BaseDirectory) 

before running it as Service E.g.

static void Main(string[] args)
        {
            if (args.Length == 0)
            {
                Directory.SetCurrentDirectory(AppDomain.CurrentDomain.BaseDirectory);
                RunAsService();
            }
            else
            {
                RunAsConsole();
            }
        }

---

You need to find the path to your service's assembly, like this:

static readonly string assemblyPath = 
    Path.GetDirectoryName(typeof(MyClass).Assembly.Location);

using (FileStream stream = File.OpenRead(Path.Combine(assemblyPath, "MyService.xml"))

---

When a Windows Service is launched, the current directory is the system directory, as you indeed seem to be finding. It is the current directory that is used to resolve relative paths into absolute paths, not your application (service) directory. (Check the Environment.CurrentDirectory variable if you want to confirm this.)

The following helper method may come in handy here:

public static string GetAppRelativePath(string path)
{
    return Path.Combine(Path.GetDirectoryName(
        Assembly.GetEntryAssembly().Location), path);
}

Which you can then use as:

using (FileStream stream = new FileStream(Utilities.GetAppRelativePath(
    "MyService.xml"), FileMode.Open, FileAccess.Read))
{
    // Read file
}

The path will then resolve to C:\Services\MyService\MyService.xml, as you want.