C# – How to get the path of the assembly the code is in

cmbunitnetreflection

Is there a way to get the path for the assembly in which the current code resides? I do not want the path of the calling assembly, just the one containing the code.

Basically my unit test needs to read some xml test files which are located relative to the dll. I want the path to always resolve correctly regardless of whether the testing dll is run from TestDriven.NET, the MbUnit GUI or something else.

Edit: People seem to be misunderstanding what I'm asking.

My test library is located in say

C:\projects\myapplication\daotests\bin\Debug\daotests.dll

and I would like to get this path:

C:\projects\myapplication\daotests\bin\Debug\

The three suggestions so far fail me when I run from the MbUnit Gui:

  • Environment.CurrentDirectory
    gives c:\Program Files\MbUnit

  • System.Reflection.Assembly.GetAssembly(typeof(DaoTests)).Location
    gives C:\Documents and
    Settings\george\Local
    Settings\Temp\ ….\DaoTests.dll

  • System.Reflection.Assembly.GetExecutingAssembly().Location
    gives the same as the previous.

Best Answer

I've defined the following property as we use this often in unit testing.

public static string AssemblyDirectory
{
    get
    {
        string codeBase = Assembly.GetExecutingAssembly().CodeBase;
        UriBuilder uri = new UriBuilder(codeBase);
        string path = Uri.UnescapeDataString(uri.Path);
        return Path.GetDirectoryName(path);
    }
}

The Assembly.Location property sometimes gives you some funny results when using NUnit (where assemblies run from a temporary folder), so I prefer to use CodeBase which gives you the path in URI format, then UriBuild.UnescapeDataString removes the File:// at the beginning, and GetDirectoryName changes it to the normal windows format.